Proving properties of an isosceles trapezoid
An isosceles trapezoid is defined by its two legs being equal in length. Why should that force its base angles, and its two diagonals, to be equal as well?
Where you are 4 in Proofs about parallelograms
When you reach for it
A proof or a problem has a trapezoid in it whose two legs are the same length, and it needs one of the facts that follow: the two angles at one base are equal, or the two diagonals are equal. Or it runs the other way, and the job is to show that a trapezoid is isosceles because its base angles or its diagonals match. The definition gives none of this. It speaks only of the legs, and the equal angles and equal diagonals have to be earned, the way the equal sides of a parallelogram were earned in proving opposite sides congruent.
The words, exactly. As in trapezoids and what a definition does, a trapezoid here is a quadrilateral with at least one pair of parallel sides. The pair of parallel sides we use are its bases, and the other two sides are its legs. The two angles at the ends of one base are a pair of base angles. So a trapezoid ABCD with \overline{AB} \parallel \overline{DC} has two pairs: \angle D and \angle C at the base \overline{DC}, and \angle A and \angle B at the base \overline{AB}. An isosceles trapezoid is a trapezoid whose legs are congruent.
Why one more hypothesis is needed. With “at least one pair”, that definition lets in more than it seems to. Every parallelogram is a trapezoid, and if one pair of its opposite sides is called the bases, its legs are congruent by the Opposite Sides Theorem. But a parallelogram’s base angles are supplementary, not equal, unless it is a rectangle; Figure 7 in the Boundaries has one with base angles 60^\circ and 120^\circ. So the theorems below cannot rest on congruent legs alone. Every one of them assumes all three of these about ABCD:
- (H1) the bases are parallel: \overline{AB} \parallel \overline{DC};
- (H2) the legs are congruent: \overline{AD} \cong \overline{BC};
- (H3) the legs are not parallel: \overline{AD} \nparallel \overline{BC}.
In this lesson, “isosceles trapezoid” means a quadrilateral with all three.
What H3 says about the bases. Given H1, the legs are parallel exactly when the bases are the same length. If AB = DC as well as \overline{AB} \parallel \overline{DC}, then ABCD is a parallelogram, since one pair of opposite sides both parallel and congruent is enough (proving a quadrilateral is a parallelogram), and its legs are parallel. If the legs are parallel, ABCD is a parallelogram by definition and AB = DC by the Opposite Sides Theorem. So H3 says the bases have different lengths. Name the figure so that \overline{DC} is the longer one, as in Figure 1.
Other books. A book that defines a trapezoid as having exactly one pair of parallel sides has H3 built in, so for it every trapezoid with congruent legs satisfies all three. A book that defines an isosceles trapezoid as a trapezoid with a pair of congruent base angles lets in rectangles and shuts out every other parallelogram. Each choice moves the same facts between definition and theorem, which is why a proof should say which one it uses.
The procedure
Theorem 1: the base angles
Given. ABCD satisfies H1, H2 and H3, with \overline{DC} the longer base.
Prove. \angle ADC \cong \angle BCD and \angle DAB \cong \angle CBA.
Plan. The two angles at D and C sit at opposite ends of the figure, and no pair of triangles in Figure 1 holds them together with the legs. So make some. From A and B drop perpendiculars to \overleftrightarrow{DC}, meeting it at P and Q. These are auxiliary lines, and they cut off two right triangles, \triangle APD and \triangle BQC, whose hypotenuses are the two legs. The two altitudes are equal because parallel lines stay the same distance apart, so the right triangles match by hypotenuse leg.
| Statements | Reasons |
|---|---|
| 1. ABCD satisfies H1, H2 and H3 | given |
| 2. \overline{AP} \perp \overleftrightarrow{DC} at P and \overline{BQ} \perp \overleftrightarrow{DC} at Q | through a point off a line there is exactly one perpendicular to it |
| 3. \overline{AP} \parallel \overline{BQ} | two lines perpendicular to one line are parallel: line 2 |
| 4. ABQP is a parallelogram | definition of parallelogram: \overline{AB} \parallel \overline{PQ} (H1) and line 3 |
| 5. \overline{AP} \cong \overline{BQ} | Opposite Sides Theorem: line 4 |
| 6. \angle APD and \angle BQC are right angles | definition of perpendicular: line 2 |
| 7. \triangle APD \cong \triangle BQC | HL: hypotenuses \overline{AD} \cong \overline{BC} (H2), legs \overline{AP} \cong \overline{BQ} (line 5), line 6 (P \ne D and Q \ne C, shown below) |
| 8. \angle ADP \cong \angle BCQ | corresponding parts of congruent triangles: line 7 |
| 9. \angle ADC \cong \angle BCD | line 8, because P lies on \overrightarrow{DC} and Q on \overrightarrow{CD} (shown below) |
| 10. \angle DAB is supplementary to \angle ADC, and \angle CBA to \angle BCD | Same Side Interior Angles Theorem: H1, transversals \overline{AD} and \overline{BC} |
| 11. \angle DAB \cong \angle CBA | supplements of congruent angles are congruent: lines 9 and 10 |
Lines 3 to 5. Each altitude makes a right angle with \overleftrightarrow{DC}, and those two right angles are a corresponding pair, so the altitudes are parallel by the converse of the Corresponding Angles Theorem. \overline{AB} lies on a line parallel to \overleftrightarrow{DC}, and \overline{PQ} lies on \overleftrightarrow{DC}, so both pairs of opposite sides of ABQP are parallel. That makes AP = BQ, and it also makes PQ = AB, which the next paragraph needs.
Line 7, and the order of the letters. In \triangle APD the right angle is at P, so the hypotenuse is \overline{AD}, one leg of the trapezoid; in \triangle BQC it is \overline{BC}, the other. The pairing is A \leftrightarrow B, P \leftrightarrow Q, D \leftrightarrow C, so the angle at D matches the angle at C and \overline{DP} matches \overline{CQ}.
Line 9: where the feet land. Line 8 is about \angle ADP, the angle between the leg and the ray from D toward P. It is the trapezoid’s angle \angle ADC only if P lies on the ray \overrightarrow{DC}; if P fell on the other side of D, \angle ADP would be the supplement of \angle ADC. Where P falls depends on lengths, not on the look of the figure (what a diagram may assume), so it has to be proved. Two things are used: PQ = AB, and Q lies on the C side of P, because \overline{PQ} runs the same way as \overline{AB}, which runs the same way as \overline{DC}.
- P is not D. If it were, \overline{AD} would be the perpendicular \overline{AP}, so BC = AD = AP = BQ. But the perpendicular \overline{BQ} is the shortest segment from B to the line (distance from a point to a line), so the only segment from B to the line as short as it is \overline{BQ} itself, and C = Q. Then both legs would be perpendicular to \overline{DC}, hence parallel, against H3. In the same way Q is not C.
- P is not beyond D. Suppose D lay between P and C. Then PC = PD + DC, which is more than AB, so Q, which is AB along from P toward C, lies between P and C and PC = PQ + QC = AB + QC. Line 7 gives PD = QC. Taking that from both expressions for PC leaves DC = AB, which is false because \overline{DC} is the longer base.
So P lies on \overrightarrow{DC}, and the same two arguments from the other end put Q on \overrightarrow{CD}. The four points then come in the order D, P, Q, C, and the long base is DC = DP + PQ + QC with PQ = AB and DP = QC. That gives a formula the examples use:
Line 10. Each leg is a transversal of the two bases, so the two angles along one leg are same side interior angles and add to 180^\circ. The angles at A and B are supplements of congruent angles, so they are congruent too (supplements of congruent angles).
Theorem 2: the diagonals
Given. ABCD satisfies H1, H2 and H3.
Prove. \overline{AC} \cong \overline{BD}.
Plan. Each diagonal is a side of a triangle standing on the long base: \overline{AC} of \triangle ADC and \overline{BD} of \triangle BCD. The two triangles overlap (proofs with overlapping triangles), share \overline{DC}, have the congruent legs as sides, and have the base angles of Theorem 1 between those sides. That is SAS.
| Statements | Reasons |
|---|---|
| 1. ABCD satisfies H1, H2 and H3 | given |
| 2. \overline{AD} \cong \overline{BC} | H2: line 1 |
| 3. \angle ADC \cong \angle BCD | Theorem 1: line 1 |
| 4. \overline{DC} \cong \overline{CD} | reflexive property of congruence |
| 5. \triangle ADC \cong \triangle BCD | SAS: lines 2, 3 and 4 |
| 6. \overline{AC} \cong \overline{BD} | corresponding parts of congruent triangles: line 5 |
In \triangle ADC the angle at D lies between \overline{DA} and \overline{DC}, and in \triangle BCD the angle at C lies between \overline{CB} and \overline{CD}, so each angle is between the two sides used, as SAS needs. The pairing is A \leftrightarrow B, D \leftrightarrow C, C \leftrightarrow D, which sends \overline{AC} to \overline{BD}.
The converses, and what each one needs
Both theorems can be run backwards, and both converses are true, but with different hypotheses. Examples 5 and 6 prove them.
| Statement | Assumes | Concludes |
|---|---|---|
| Theorem 1 | H1, H2, H3 | \angle D \cong \angle C and \angle A \cong \angle B |
| Theorem 2 | H1, H2, H3 | \overline{AC} \cong \overline{BD} |
| Converse 1 | H1 and \angle D \cong \angle C | \overline{AD} \cong \overline{BC} |
| Converse 2 | H1 and \overline{AC} \cong \overline{BD} | \overline{AD} \cong \overline{BC} |
The converses do not need H3, and they do not deliver it either. If the legs of such a figure are parallel it is a parallelogram with a pair of equal base angles or equal diagonals, which forces a rectangle (proving a parallelogram is a rectangle). So each converse says: a trapezoid with a pair of congruent base angles, or with congruent diagonals, is an isosceles trapezoid or a rectangle.
Worked examples
Four angles from one unknown
ABCD is an isosceles trapezoid (H1, H2 and H3) with m\angle D = (3x + 10)^\circ and m\angle C = (5x - 22)^\circ. Find x and all four angles.
Solution\angle D and \angle C are the base angles at \overline{DC}, so Theorem 1 makes them congruent and their measures equal:
So m\angle D = 3(16) + 10 = 58^\circ, and m\angle C = 5(16) - 22 = 58^\circ agrees. The leg \overline{AD} is a transversal of the bases, so \angle A and \angle D are same side interior angles: m\angle A = 180^\circ - 58^\circ = 122^\circ. Theorem 1 gives m\angle B = 122^\circ as well.
Check. 58 + 58 + 122 + 122 = 360, the angle sum of a quadrilateral.
Diagonals and legs in one unknown
ABCD is an isosceles trapezoid with AC = 4y - 5, BD = 2y + 9 and AD = y + 3. Find y, the length of each diagonal, and BC.
SolutionBy Theorem 2 the diagonals are congruent, so
Then AC = 4(7) - 5 = 23 and BD = 2(7) + 9 = 23. The legs are congruent by H2, so BC = AD = 7 + 3 = 10.
Height, area and a diagonal from the sides
ABCD is an isosceles trapezoid with bases AB = 8 and DC = 20 and legs 10, and P and Q are the feet of the altitudes from A and B, as in Figure 2. Find DP, the height AP, the area, and the length of each diagonal.
SolutionThe proof of Theorem 1 showed that the feet P and Q land between D and C with \displaystyle DP = QC = \frac{DC - AB}{2}, so
\triangle APD has a right angle at P and hypotenuse AD = 10, so by the Pythagorean theorem
The area of a trapezoid is half the sum of the bases times the height (area of a trapezoid):
For the diagonal \overline{AC}, use the right triangle \triangle APC. Its leg along the base is PC = DC - DP = 20 - 6 = 14, so
By Theorem 2, \displaystyle BD = 2\sqrt{65} as well.
Check. From the other side, \triangle BQD has legs BQ = 8 and QD = 20 - 6 = 14, giving the same \displaystyle \sqrt{260}.
Where the diagonals cross
Given. ABCD is an isosceles trapezoid (H1, H2 and H3) whose diagonals meet at E, as in Figure 4.
Prove. \overline{AE} \cong \overline{BE} and \overline{CE} \cong \overline{DE}.
Plan. \overline{AE} and \overline{BE} are two sides of \triangle ABE, so they are congruent if the angles opposite them are, by the converse of the Base Angles Theorem. Those angles, \angle BAE and \angle ABE, are angles of the overlapping triangles \triangle CBA and \triangle DAB, which stand on the short base. Then the rest of each diagonal is what is left over.
| Statements | Reasons |
|---|---|
| 1. ABCD satisfies H1, H2 and H3; the diagonals meet at E | given |
| 2. \overline{AD} \cong \overline{BC} | H2: line 1 |
| 3. \angle DAB \cong \angle CBA | Theorem 1: line 1 |
| 4. \overline{AB} \cong \overline{BA} | reflexive property of congruence |
| 5. \triangle DAB \cong \triangle CBA | SAS: lines 2, 3 and 4 |
| 6. \angle ABD \cong \angle BAC | corresponding parts of congruent triangles: line 5 |
| 7. \overline{AE} \cong \overline{BE} | converse of the Base Angles Theorem in \triangle ABE: line 6 |
| 8. AC = BD | Theorem 2: line 1 |
| 9. {CE = AC - AE} | Segment Addition Postulate: E is between A and C |
| 10. {DE = BD - BE} | Segment Addition Postulate: E is between B and D |
| 11. \overline{CE} \cong \overline{DE} | lines 7 to 10: equal lengths minus equal lengths |
Line 5 pairs D \leftrightarrow C, A \leftrightarrow B, B \leftrightarrow A. The angle at B in \triangle DAB, between \overline{BA} and \overline{BD}, matches the angle at A in \triangle CBA, between \overline{AB} and \overline{AC}: that is line 6. E lies on \overline{BD} and on \overline{AC}, so \angle ABD is \angle ABE and \angle BAC is \angle BAE, the two angles of \triangle ABE at its base \overline{AB}. Lines 9 and 10 use the fact that the diagonals of a quadrilateral like this one cross inside it, a position fact the figure may be trusted for.
The pieces match in pairs, AE = BE and CE = DE, but E is not the midpoint of either diagonal. If it were the midpoint of one, say AE = CE, then with AE = BE and CE = DE all four pieces would be equal, the diagonals would bisect each other, and ABCD would be a parallelogram by the Diagonals Converse (proving a quadrilateral is a parallelogram), against H3.
Converse 1: equal base angles give equal legs
Given. \overline{AB} \parallel \overline{DC} and \angle ADC \cong \angle BCD, as in Figure 5.
Prove. \overline{AD} \cong \overline{BC}.
Plan. Drop the same two altitudes as in Theorem 1. The right triangles now have a matching angle and a matching leg, which is AAS. First, though, the feet must land on the correct side of D and C, and that depends on the size of the given angles.
The size of the angles. If \angle D and \angle C are right angles, both legs are perpendicular to \overline{DC}, so they are parallel and ABCD is a parallelogram with a right angle, a rectangle, and AD = BC by the Opposite Sides Theorem. If they are obtuse, their supplements \angle A and \angle B are acute and congruent, and the whole argument below can be run from the base \overline{AB} instead, dropping perpendiculars from D and C. So it is enough to treat \angle D and \angle C acute.
Where the feet land. When \angle ADC is acute, the foot P lies on \overrightarrow{DC} and is not D. If P were D, \angle ADC would be a right angle. If P lay on the opposite ray, the angle of \triangle APD at D would be the supplement of \angle ADC, which is obtuse, and a triangle with a right angle and an obtuse angle would have angles adding to more than 180^\circ (triangle angle sum). In the same way Q lies on \overrightarrow{CD}.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \parallel \overline{DC}; \angle ADC \cong \angle BCD, both acute | given (acute case) |
| 2. \overline{AP} \perp \overleftrightarrow{DC} at P and \overline{BQ} \perp \overleftrightarrow{DC} at Q | exactly one perpendicular from a point to a line |
| 3. \overline{AP} \cong \overline{BQ} | ABQP is a parallelogram, as in lines 3 to 5 of Theorem 1 |
| 4. \angle APD \cong \angle BQC | all right angles are congruent: line 2 |
| 5. \angle ADP \cong \angle BCQ | line 1, because P is on \overrightarrow{DC} and Q on \overrightarrow{CD} |
| 6. \triangle APD \cong \triangle BQC | AAS: lines 5, 4 and 3 |
| 7. \overline{AD} \cong \overline{BC} | corresponding parts of congruent triangles: line 6 |
In \triangle APD the side \overline{AP} joins A and P, so it is not between the angles at D and P; it lies opposite the angle at D. That is why line 6 is AAS and not ASA.
So the legs are congruent. They are parallel only if \angle D and \angle C, a same side interior pair along \overline{DC}, add to 180^\circ, and being equal, that means both are right angles: the rectangle case.
Converse 2: equal diagonals give equal legs
Given. \overline{AB} \parallel \overline{DC} and \overline{AC} \cong \overline{BD}, as in Figure 6.
Prove. \overline{AD} \cong \overline{BC}.
Plan. The triangles \triangle ACD and \triangle BDC have the equal diagonals and the shared base as sides, so SAS would finish if the angles between them, \angle ACD and \angle BDC, were known equal. Those angles sit at different ends. Bring them together with an auxiliary line: through B draw the line parallel to \overline{AC}, meeting \overleftrightarrow{DC} at E.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \parallel \overline{DC} and \overline{AC} \cong \overline{BD} | given |
| 2. The line through B parallel to \overline{AC} meets \overleftrightarrow{DC} at E | parallel postulate; it meets \overleftrightarrow{DC}, since a line parallel to \overline{AC} cannot also be parallel to \overleftrightarrow{DC}, which crosses \overline{AC} at C |
| 3. ABEC is a parallelogram | definition of parallelogram: \overline{AB} \parallel \overline{CE} (line 1) and \overline{AC} \parallel \overline{BE} (line 2) |
| 4. \overline{BE} \cong \overline{AC} | Opposite Sides Theorem: line 3 |
| 5. \overline{BE} \cong \overline{BD} | transitive property of congruence: lines 1 and 4 |
| 6. \angle BDE \cong \angle BED | Base Angles Theorem in \triangle BDE: line 5 |
| 7. \angle ACD \cong \angle BED | Corresponding Angles Theorem: \overline{AC} \parallel \overline{BE}, transversal \overleftrightarrow{DE} |
| 8. \angle BDC \cong \angle ACD | lines 6 and 7; \angle BDE is \angle BDC |
| 9. \overline{DC} \cong \overline{CD} | reflexive property of congruence |
| 10. \triangle ACD \cong \triangle BDC | SAS: lines 1, 8 and 9 |
| 11. \overline{AD} \cong \overline{BC} | corresponding parts of congruent triangles: line 10 |
Where E lands. CE = AB, and \overline{CE} runs the same way as \overline{AB}, which is the way from D to C. So E lies on the ray from D through C, beyond C. That puts C between D and E, so the ray \overrightarrow{DE} is \overrightarrow{DC} and \angle BDE is \angle BDC (line 8). It also puts \angle ACD and \angle BED in matching positions at the two crossings of \overleftrightarrow{DE}: each opens toward D, on the same side of the line, which makes them corresponding angles (line 7).
Line 10 pairs A \leftrightarrow B, C \leftrightarrow D, D \leftrightarrow C: the diagonal \overline{AC} with \overline{BD}, the angle at C with the angle at D, and \overline{CD} with \overline{DC}. Then \overline{AD} matches \overline{BC}. Nothing in this proof used which base is longer, and nothing used H3: the argument works for a rectangle as well.
Boundaries
The legs must not be parallel. Figure 7 is a parallelogram with base angles 60^\circ and 120^\circ. With \overline{AB} and \overline{DC} as its bases it satisfies H1 and H2, since its legs are opposite sides, but not H3, and both theorems fail for it: its base angles differ, and its diagonals differ too, the one joining the two 60^\circ corners being the longer.
The proof of Theorem 1 shows exactly where it breaks. Lines 2 to 8 still hold (only line 1’s claim of H3 is false), so HL still gives \angle ADP \cong \angle BCQ. But the bases are now equal, the argument that kept each foot on the near side of its vertex needed DC longer than AB, and here the foot Q lands beyond C. So \angle BCQ is the supplement of \angle BCD, and line 9 fails: the proof shows only that \angle D is congruent to the supplement of \angle C, which every parallelogram already knew.
Rectangles satisfy the conclusions without the hypotheses. A rectangle fails H3, yet all four of its angles are right angles and its diagonals are equal (properties of a rectangle). H1 to H3 are enough for the theorems, not necessary for their conclusions.
Without parallel bases, nothing follows. Equal legs with no parallel sides give no equal angles: the quadrilateral with A(0, 4), B(3.6, 3.2), C(6, 0) and D(0, 0) has AD = BC = 4, yet m\angle D = 90^\circ and m\angle C \approx 53.13^\circ. Equal diagonals alone do not make a trapezoid either (Exercise 20). Both converses need H1.
Base angles, not opposite angles. Theorem 1 pairs the two angles at one base. The opposite angles \angle A and \angle C are supplementary, because \angle A is supplementary to \angle D and \angle D \cong \angle C; they are congruent only when both are right angles, which H3 rules out.
Where this goes next
Exercise 21 shows that the line through the midpoints of the two bases is perpendicular to both and is a line of symmetry: reflecting in it swaps A with B and D with C. Every fact in this lesson can be read off that one reflection, which is how a transformational proof would argue, and it is the same picture as the single line of symmetry of an isosceles triangle (proving base angles congruent) and of a kite.
Because the opposite angles of an isosceles trapezoid are supplementary, it is always a cyclic quadrilateral: one circle passes through all four corners. The altitude picture of Example 3 is also the start of the area of a trapezoid, and the segment joining the midpoints of the legs is the midsegment of a trapezoid. Where the isosceles trapezoid sits among the other four-sided figures, and why rectangles keep appearing beside it, is the subject of the quadrilateral hierarchy. In coordinates, placing the bases on horizontal lines symmetric about the y-axis turns Theorems 1 and 2 into one-line coordinate proofs.
Exercises
In every exercise, ABCD is an isosceles trapezoid in the sense of this lesson (H1, H2 and H3: \overline{AB} \parallel \overline{DC}, \overline{AD} \cong \overline{BC}, legs not parallel), with \overline{DC} the longer base and P and Q the feet of the perpendiculars from A and B to \overline{DC}, unless the exercise says otherwise.
Using the theorems
- m\angle D = 71^\circ. Find the other three angles.
- m\angle A = 124^\circ. Find m\angle B, m\angle C and m\angle D.
- m\angle D = (2x + 15)^\circ and m\angle C = (4x - 25)^\circ. Find x and all four angles.
- m\angle A = (3x)^\circ and m\angle D = (x + 20)^\circ. Find x and all four angles.
- AC = 3x + 4 and BD = 5x - 10. Find x and AC.
- AD = 2y + 1, BC = 13 - y and AC = 3y + 5. Find y, the length of each leg, and BD.
- The diagonals meet at E, with AE = 5 and CE = 9. Find BE, DE and BD.
Altitudes and lengths
- AB = 10, DC = 22 and each leg is 10. Find DP, the height and the area.
- AB = 7, DC = 17 and the height is 12. Find the length of each leg and the perimeter.
- AB = 11, DC = 21 and each leg is 13. Find the height and the length of each diagonal.
- AB = 6, DC = 14 and m\angle D = 45^\circ. Find the height, the length of each leg and the area.
- DC = 30, each leg is 17 and the height is 15. Find AB.
Coordinates
- A(1, 4), B(5, 4), C(7, 0) and D(-1, 0). Show that ABCD satisfies H1, H2 and H3, and find the length of each diagonal.
- D(0, 0), C(12, 0) and A(3, 5) are given, and B is to be placed on the line y = 5 with BC = AD. Find both possible points B, and say which one makes ABCD an isosceles trapezoid.
Short proofs
Write each as a two-column proof or a paragraph.
- Prove. \angle DAC \cong \angle CBD.
- The diagonals meet at E. Prove. \triangle ADE \cong \triangle BCE.
- N is the midpoint of \overline{DC}. Prove. \overline{NA} \cong \overline{NB}.
- In this exercise only H1 is given about ABCD, together with \angle DAB \cong \angle CBA, both obtuse. Prove. \overline{AD} \cong \overline{BC}.
Reasoning
- In Figure 7, a parallelogram, the proof of Theorem 1 is attempted with P and Q the feet of the perpendiculars from A and B. Say where Q lands and which line of the proof fails.
- A(0, 2), B(3, 0), C(0, -8) and D(-7, 0). Show that the diagonals of ABCD are congruent but that no two of its sides are parallel. What does this say about Converse 2?
- M is the midpoint of \overline{AB} and N the midpoint of \overline{DC}. Explain why \overleftrightarrow{MN} is perpendicular to both bases and is a line of symmetry of ABCD.
- Explain why \angle A and \angle C are supplementary, and why they cannot be congruent.
- A book defines an isosceles trapezoid as “a trapezoid with congruent diagonals”, using the inclusive definition of trapezoid. Which quadrilaterals does that admit, compared with H1, H2 and H3?
Answers
- m\angle C = 71^\circ, m\angle A = m\angle B = 109^\circ. Theorem 1 gives \angle C \cong \angle D, and \angle A is supplementary to \angle D along the leg \overline{AD}: 180 - 71 = 109.
- m\angle B = 124^\circ, m\angle C = m\angle D = 56^\circ. \angle B \cong \angle A by Theorem 1, and \angle D is the supplement of \angle A: 180 - 124 = 56.
- x = 20; m\angle D = m\angle C = 55^\circ and m\angle A = m\angle B = 125^\circ. The base angles are equal: 2x + 15 = 4x - 25, so 40 = 2x. Then 2(20) + 15 = 55, and the angles at the other base are 180 - 55 = 125.
- x = 40; m\angle A = m\angle B = 120^\circ and m\angle C = m\angle D = 60^\circ. \angle A and \angle D lie along one leg, so they are supplementary: 3x + x + 20 = 180, giving 4x = 160. Then m\angle A = 120^\circ and m\angle D = 60^\circ, and Theorem 1 copies each to its partner.
- x = 7 and AC = 25. By Theorem 2, 3x + 4 = 5x - 10, so 14 = 2x. Then AC = 3(7) + 4 = 25, and BD = 5(7) - 10 = 25 agrees.
- y = 4, each leg is 9, and BD = 17. The legs are congruent: 2y + 1 = 13 - y, so 3y = 12. Then AD = 9 and BC = 13 - 4 = 9. By Theorem 2, BD = AC = 3(4) + 5 = 17.
- BE = 5, DE = 9 and BD = 14. Example 4 gives BE = AE and DE = CE, and then BD = BE + ED = 14, which equals AC = 5 + 9 as Theorem 2 requires.
- DP = 6, the height is 8 and the area is 128. \displaystyle DP = \frac{22 - 10}{2} = 6, then \displaystyle AP = \sqrt{10^2 - 6^2} = 8, and the area is \displaystyle \frac{10 + 22}{2} \cdot 8 = 128.
- Each leg is 13 and the perimeter is 50. \displaystyle DP = \frac{17 - 7}{2} = 5, and the leg is the hypotenuse of \triangle APD: \displaystyle \sqrt{12^2 + 5^2} = 13. Perimeter: 7 + 17 + 13 + 13 = 50.
- The height is 12 and each diagonal is 20. \displaystyle DP = \frac{21 - 11}{2} = 5, so \displaystyle AP = \sqrt{13^2 - 5^2} = 12. Then PC = 21 - 5 = 16 and \displaystyle AC = \sqrt{12^2 + 16^2} = 20; BD = 20 by Theorem 2.
- The height is 4, each leg is \displaystyle 4\sqrt{2} and the area is 40. \displaystyle DP = \frac{14 - 6}{2} = 4. \triangle APD has a right angle at P and 45^\circ at D, so it is isosceles and AP = DP = 4. Then \displaystyle AD = \sqrt{4^2 + 4^2} = 4\sqrt{2}, and the area is \displaystyle \frac{6 + 14}{2} \cdot 4 = 40.
- AB = 14. \displaystyle DP = \sqrt{17^2 - 15^2} = 8, and \displaystyle DP = \frac{DC - AB}{2} gives AB = 30 - 2(8) = 14.
- Each diagonal is \displaystyle 2\sqrt{13}. H1: \overline{AB} and \overline{DC} are both horizontal. H2: \displaystyle AD = \sqrt{2^2 + 4^2} = \sqrt{20} and \displaystyle BC = \sqrt{2^2 + 4^2} = \sqrt{20}. H3: the slope of \overline{AD} is \displaystyle \frac{4}{2} = 2 and the slope of \overline{BC} is \displaystyle \frac{-4}{2} = -2. Then \displaystyle AC = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13} and \displaystyle BD = \sqrt{6^2 + 4^2} = 2\sqrt{13}.
- B(9, 5) or B(15, 5), and only B(9, 5) gives an isosceles trapezoid. \displaystyle AD = \sqrt{3^2 + 5^2} = \sqrt{34}, and B = (x, 5) needs (12 - x)^2 + 5^2 = 34, so 12 - x = \pm 3. With B(15, 5), AB = 12 = DC and both legs have slope \displaystyle \frac{5}{3}, so ABCD is a parallelogram and fails H3.
- In the proof of Theorem 2, \triangle ADC \cong \triangle BCD with A \leftrightarrow B, D \leftrightarrow C, C \leftrightarrow D. The angle at A in \triangle ADC is \angle DAC, and it corresponds to the angle at B in \triangle BCD, which is \angle CBD. So \angle DAC \cong \angle CBD by corresponding parts.
- \overline{AD} \cong \overline{BC} by H2, and Example 4 gives \overline{AE} \cong \overline{BE} and \overline{DE} \cong \overline{CE}. So \triangle ADE \cong \triangle BCE by SSS, pairing A \leftrightarrow B, D \leftrightarrow C, E \leftrightarrow E.
- \overline{AD} \cong \overline{BC} by H2, \angle ADN \cong \angle BCN because they are the base angles \angle ADC and \angle BCD of Theorem 1 (N is on \overline{DC}), and \overline{DN} \cong \overline{CN} because N is the midpoint. Each angle lies between the two sides named, so \triangle ADN \cong \triangle BCN by SAS, and corresponding parts give \overline{NA} \cong \overline{NB}.
- \angle D is supplementary to \angle A and \angle C to \angle B (same side interior angles along the legs, by H1), so \angle ADC \cong \angle BCD as supplements of congruent angles, and they are acute because \angle A and \angle B are obtuse. Converse 1 (Example 5, acute case) then gives \overline{AD} \cong \overline{BC}.
- Q lands beyond C, on the ray opposite \overrightarrow{CD}. Lines 2 to 8 still hold, so \angle ADP \cong \angle BCQ, but \angle BCQ is now the supplement of \angle BCD, not \angle BCD itself. Line 9 fails. The step that kept the feet on the near side used DC > AB, and in a parallelogram DC = AB.
- AC = 10 and BD = 10, since \overline{AC} runs from y = 2 to y = -8 on the y-axis and \overline{BD} from x = 3 to x = -7 on the x-axis. The slopes are \displaystyle -\frac{2}{3} for \overline{AB}, \displaystyle -\frac{8}{7} for \overline{DC}, \displaystyle \frac{2}{7} for \overline{AD} and \displaystyle \frac{8}{3} for \overline{BC}, so neither pair of opposite sides is parallel. Congruent diagonals alone do not make a trapezoid, so Converse 2 needs H1 among its hypotheses.
- By Exercise 17, N is the same distance from A as from B, so it lies on the perpendicular bisector of \overline{AB} (perpendicular bisector theorem), and M, the midpoint, lies on it too. M and N are different points, being on different parallel lines, so \overleftrightarrow{MN} is the perpendicular bisector of \overline{AB}. In the same way, triangles DAM and CBM are congruent by SAS (legs, base angles at A and B, halves of \overline{AB}), so MD = MC and \overleftrightarrow{MN} is also the perpendicular bisector of \overline{DC}. Reflection in this line sends A to B and D to C, and so carries the trapezoid onto itself.
- \angle A is supplementary to \angle D along the leg \overline{AD}, and \angle D \cong \angle C by Theorem 1, so m\angle A + m\angle C = 180^\circ. If they were also congruent, each would be 90^\circ, and then \angle D and \angle C would be right angles, both legs would be perpendicular to \overline{DC} and so parallel, which H3 forbids.
- It admits every quadrilateral satisfying H1, H2 and H3, and also every rectangle, and nothing else. Theorem 2 shows H1 to H3 give congruent diagonals, and a rectangle’s diagonals are congruent. Conversely, Converse 2 gives congruent legs; if the legs are not parallel that is H1 to H3, and if they are, the figure is a parallelogram with congruent diagonals, a rectangle. Other parallelograms are shut out, since their diagonals differ.