Medians and altitudes
From one corner of a triangle, which line to the opposite side cuts the triangle fairly in two, and which one measures how tall the triangle stands?
Where you are 4 in Special segments in a triangle
Why this exists
Pick a corner of a triangle and draw a segment from it to the side across the way. There are endlessly many such segments, one for every point of that side, and most of them are of no special interest. Two of them are.
One lands exactly halfway along the opposite side. It cuts the triangle into two pieces that usually look nothing alike, and yet always have the same area. That is the median.
The other lands so that it meets the opposite side squarely, at a right angle. It is the shortest way from the corner to that side, and so it is the honest answer to “how tall is this triangle?” That is the altitude, and it is the height in the area formula A = \tfrac12 bh. Without it, “height” has no meaning for a triangle that is leaning over.
Every triangle has three of each, one from every corner, and this lesson is about what they are, where they fall, and what they let you find.
Definition
In \triangle ABC, the side across from vertex A is \overline{BC}, and it is called the opposite side of A.
A median of a triangle is a segment from a vertex to the midpoint of the opposite side. In Figure 1, M is the midpoint of \overline{BC}, so BM = MC (the tick marks say so), and \overline{AM} is the median from A.
An altitude of a triangle is a segment from a vertex, perpendicular to the line containing the opposite side, that ends on that line. The point where it ends is its foot. In Figure 1, \overline{AD} is the altitude from A: \overline{AD} \perp \overline{BC} (the square mark says so), and D is its foot.
Three details in the definition of altitude matter later.
- It says the line containing the opposite side, not the side itself. The foot may land on the side, at one of its ends, or out on the side extended beyond the triangle. All three happen.
- When the foot lands exactly on a vertex, the altitude is one of the triangle’s own sides. In a right triangle, each leg is the altitude to the other leg.
- The word “altitude” is used for the segment and also for its length. We write h_A for the length of the altitude from A, and call it the height of the triangle on the base \overline{BC}. Every side can serve as a base, each with its own height.
A median and an altitude are both defined by where they land, and they land by different rules: one at the midpoint, the other at a right angle. From one vertex they are usually two different segments, as in Figure 1. They are also different from the two other families drawn in a triangle: a perpendicular bisector of a side meets it at its midpoint at a right angle but need not pass through any vertex, and an angle bisector halves the angle at a vertex but need not land at the midpoint or at a right angle.
Intuition
A median is a fair cut. In Figure 2 the median \overline{AM} splits \triangle ABC into \triangle ABM (shaded) and \triangle AMC. They have different shapes, but look at what they share. Their bases, \overline{BM} and \overline{MC}, are the two equal halves of \overline{BC}. Their heights are the same segment: both are measured from A straight down to line BC, which is the dashed h. Equal bases and equal heights make equal areas.
An altitude is a plumb line. Turn the triangle so that one side lies flat on the floor. The altitude to that side is a string hanging straight down from the top corner to the floor. If the triangle leans far enough, the corner hangs out past the end of its base, and the string reaches the floor outside the triangle. That is the obtuse case in Figure 3.
In an acute triangle every altitude falls inside. In a right triangle two of them are the legs, and the third runs from the right angle to the hypotenuse. In an obtuse triangle the altitude from the obtuse corner falls inside, and the other two fall outside, onto sides extended past the obtuse corner. A median never does anything like that: the midpoint is always on the side, so the median always runs through the inside of the triangle.
Results
Three of each. Every triangle has exactly three medians and exactly three altitudes, one of each from every vertex.
A median halves the area. If \overline{AM} is a median of \triangle ABC, then
The two halves have equal area but in general are not congruent.
Medians lie inside. Every point of a median other than its two ends lies inside the triangle.
Where an altitude lands. Let D be the foot of the altitude from A on line BC.
| The angles at B and C | The foot D |
|---|---|
| both acute | strictly between B and C, on the side |
| \angle B is a right angle | at B; the altitude is the side \overline{AB} |
| \angle B is obtuse | beyond B, on side \overline{CB} extended past B |
The same holds with B and C swapped. So in an acute triangle all three altitudes lie inside; in a right triangle two altitudes are the legs and the third lies inside; in an obtuse triangle the two altitudes from the acute vertices lie outside and the third lies inside.
One area, three heights. Write a = BC, b = CA and c = AB for the sides opposite A, B and C. Any side may be the base, so
and therefore a\,h_A = b\,h_B = c\,h_C. The longest side has the shortest altitude, and the shortest side the longest.
Isosceles triangles. If AB = AC, the median from A is also the altitude from A, and it bisects \angle BAC. Conversely, if the median from A is also the altitude from A, then AB = AC.
Right triangles. In a right triangle, the median to the hypotenuse is half as long as the hypotenuse. Example 6 proves it.
Where they meet. The three medians of a triangle pass through one point, the Centroid, and the lines containing the three altitudes pass through one point, the orthocenter. Each of those is a lesson of its own.
Why it is true
A median halves the area. Let \overline{AM} be the median from A and let h be the distance from A to line BC, the length of the altitude from A. Triangle ABM has base \overline{BM} on line BC and its third vertex at A, so its height is h. Triangle AMC has base \overline{MC} on the same line and the same third vertex, so its height is also h. Since BM = MC = \tfrac12 BC,
and the two together make up \triangle ABC, so each is half of it.
Medians lie inside. The midpoint M lies on \overline{BC} strictly between B and C, so it is a point of the triangle that is not a vertex. A triangle together with its inside has no dents: the segment joining a vertex to any other point of the opposite side passes through the inside. So every point of \overline{AM} between A and M is inside the triangle.
Where an altitude lands. The foot D is somewhere on line BC, and there are five places it can be: beyond B, at B, strictly between B and C, at C, or beyond C. Take them in turn.
- If D is strictly between B and C, then \triangle ABD has a right angle at D. The angles of a triangle add to 180^\circ (the triangle angle sum) and the right angle uses 90^\circ of it, so \angle ABD is acute. That angle is \angle B of the triangle, because ray BD points along ray BC. The same argument in \triangle ACD makes \angle C acute.
- If D = B, then \overline{AB} \perp \overline{BC}, so \angle B is a right angle.
- If D is beyond B, then \triangle ABD still has a right angle at D, so \angle ABD is acute. Now \angle ABD and \angle ABC form a linear pair, because D, B, C lie on one line with B between, so m\angle ABC = 180^\circ - m\angle ABD, which is more than 90^\circ. So \angle B is obtuse.
The last two places are the same with B and C swapped. The five outcomes (both acute, \angle B right, \angle B obtuse, \angle C right, \angle C obtuse) cannot happen two at a time, since a triangle has at most one angle of 90^\circ or more. So each angle condition happens exactly when its position does, and the table can be read in both directions.
Isosceles triangles. Suppose AB = AC and M is the midpoint of \overline{BC}.
| Statements | Reasons |
|---|---|
| AB = AC | given |
| BM = MC | M is the midpoint of \overline{BC} |
| AM = AM | a segment equals itself |
| \triangle ABM \cong \triangle ACM | SSS (SSS, SAS, ASA, AAS) |
| \angle AMB \cong \angle AMC | corresponding parts of congruent triangles |
| m\angle AMB + m\angle AMC = 180^\circ | they form a linear pair |
| m\angle AMB = m\angle AMC = 90^\circ | two equal measures that add to 180^\circ |
| \angle BAM \cong \angle CAM | corresponding parts of congruent triangles |
The seventh line says the median is perpendicular to \overline{BC}, so it is the altitude from A, and the last line says it bisects \angle BAC. (The Perpendicular bisectors of a triangle lesson reaches the perpendicular half of this a second way, from A being equidistant from B and C.)
For the converse, suppose the median \overline{AM} is also perpendicular to \overline{BC}. Then line AM passes through the midpoint of \overline{BC} at a right angle, so it is the perpendicular bisector of \overline{BC}. Every point on the perpendicular bisector is the same distance from B and C (Perpendicular bisectors of a triangle), and A is on it, so AB = AC.
Worked examples
A median as an expression
In \triangle ABC, \overline{AD} is the median to \overline{BC}, with BD = 3x + 2 and DC = 5x - 6. Find x and BC. If \triangle ABD has area 45, what is the area of \triangle ABC?
SolutionA median ends at the midpoint of the opposite side, so D cuts \overline{BC} into two equal pieces. That is the only fact needed for the first part:
Then BD = 3(4) + 2 = 14 and DC = 5(4) - 6 = 14, which agree as they must, so BC = 28.
A median halves the area, so \triangle ADC also has area 45, and the whole triangle has area 2 \cdot 45 = 90.
Nothing in the problem says \overline{AD} is perpendicular to \overline{BC}, so it would be wrong to call it an altitude or to use it as the height.
A median on the coordinate plane
The triangle has vertices A(6, 6), B(-2, 1) and C(8, 3). Find the endpoint and the length of the median from A, and decide whether it is also an altitude.
SolutionThe median from A ends at the midpoint of the opposite side, \overline{BC}. Average the coordinates of B and C (the midpoint formula):
Its length is the distance from A(6, 6) to M(3, 2), by the The distance formula:
To be an altitude, \overline{AM} would have to be perpendicular to \overline{BC}, and two lines that are neither horizontal nor vertical are perpendicular exactly when their slopes multiply to -1 (Slopes of perpendicular lines). The slope of \overline{BC} is \displaystyle \frac{3 - 1}{8 - (-2)} = \frac{1}{5} and the slope of \overline{AM} is \displaystyle \frac{6 - 2}{6 - 3} = \frac{4}{3}. Their product is \displaystyle \frac{4}{15}, not -1, so the median from A is not an altitude.
Three altitudes from one area
The triangle has vertices A(1, 6), B(-1, -2) and C(7, -2). Find the lengths of all three altitudes.
SolutionStart with the altitude that is easy to see. B and C have the same y-coordinate, so \overline{BC} is horizontal, and the altitude from A is vertical. Its foot is (1, -2), which lies between B and C, and its length is the drop in y:
The base is BC = 7 - (-1) = 8, so
The other two altitudes are slanted, and finding their feet would take work. The area makes that unnecessary, because the same area comes out whichever side is the base. Find the other two sides with the The distance formula:
Now \text{Area} = \tfrac12 \cdot AC \cdot h_B, so h_B = \dfrac{2 \cdot 32}{AC}, and the same for h_C:
Check. The longest side, AC = 10, has the shortest altitude, 6.4, and the shortest side, BC = 8, has the longest, 8, as the products a\,h_A = b\,h_B = c\,h_C = 64 require.
An altitude that falls outside
The triangle has vertices A(-3, 4), B(0, 0) and C(6, 0). Find the foot of the altitude from A and say whether it lies on side \overline{BC}. Then find the area of the triangle and the length of the altitude from C.
Side \overline{BC} lies on the x-axis, so the altitude from A is vertical and its foot is directly below A, at D(-3, 0). Side \overline{BC} runs only from x = 0 to x = 6, so D is not on it: it is 3 units beyond B, on side \overline{CB} extended. By the table in Results, that means \angle B is obtuse, and the picture agrees.
The altitude still measures the height. Its length is AD = 4, and the base is the side itself, BC = 6, not DC:
For the altitude from C, use the area again with \overline{AB} as the base. By the The distance formula, \displaystyle AB = \sqrt{3^2 + 4^2} = 5, so
Since \angle B is obtuse, the table (with the roles of the vertices changed) says this altitude also lands outside, on side \overline{AB} extended past B. Its length does not care where it lands.
The median from the apex of an isosceles triangle
In \triangle ABC, AB = AC = 13 and BC = 10, as in Figure 5. Find the length of the median \overline{AM}, the area of the triangle, and the length of the altitude from B.
M is the midpoint of \overline{BC}, so BM = 5. Because AB = AC, the median from A is also the altitude from A (Results), so \triangle ABM has a right angle at M with hypotenuse AB = 13. By the The Pythagorean theorem,
The median is the height on base \overline{BC}, so
With \overline{AC} as the base instead,
The Pythagorean step was allowed only because the triangle is isosceles. In a triangle with AB \ne AC, the median is not perpendicular to the base and \triangle ABM is not a right triangle.
The median to the hypotenuse
Show that in every right triangle the median to the hypotenuse is half as long as the hypotenuse. Then find that median, and the altitude to the hypotenuse, when the legs are 5 and 12.
SolutionPlace the triangle on a grid so that the right angle is at the origin, C(0, 0), with the legs along the axes: A(2p, 0) and B(0, 2q), where p and q are positive. Writing the coordinates as 2p and 2q is only to keep the midpoint free of fractions; any right triangle can be placed this way.
The hypotenuse is \overline{AB} and its midpoint is M(p, q). By the The distance formula,
So CM = \tfrac12 AB. Since M is the midpoint, MA and MB are also \tfrac12 AB: the midpoint of the hypotenuse is the same distance from all three vertices, which makes it the circumcenter of a right triangle.
With legs 5 and 12, the hypotenuse is \displaystyle \sqrt{5^2 + 12^2} = 13, and the median to it is \tfrac{13}{2} = 6.5. The altitude to the hypotenuse comes from the area, which is \tfrac12 \cdot 5 \cdot 12 = 30, using one leg as the base and the other as its height:
The altitude is shorter than the median, as it must be: the altitude is the shortest segment from C to the line AB. They would be the same segment only if the legs were equal.
A median of a median
\triangle ABC has area 48. \overline{AM} is a median, and N is the midpoint of \overline{AM}. Find the area of \triangle ABN.
SolutionThe median \overline{AM} halves the area, so \triangle ABM has area 24.
Now look at \triangle ABM on its own. N is the midpoint of its side \overline{AM}, so \overline{BN} is a median of \triangle ABM, and it halves that triangle’s area:
which is a quarter of the original triangle. The fact was used twice, in two different triangles. That is typical: a median of one triangle is often a side of a smaller one, and the half-area fact applies afresh each time.
Boundaries
An altitude lands on a line, not always on the side. In an obtuse triangle two of the feet are outside the triangle, on sides extended. The length of the altitude is still the height for that base, and the base in the area formula is still the side itself, not the side plus its extension.
“Altitude” means a segment, a length, or a line. The segment is what the definition gives. The length is what goes into the area formula. When altitudes are said to meet at one point, it is the lines that meet: in an obtuse triangle the three altitude segments do not touch each other at all.
Medians and altitudes are different in general. From the same vertex they coincide exactly when the two sides from that vertex are equal. Using a median as a height, or putting an altitude’s foot at the midpoint, is only allowed in that case.
Equal area is not congruence. The two halves made by a median have the same area, but they are congruent only when the two sides from the vertex are equal.
A midpoint or a right angle must be given, not seen. A segment that looks as if it lands halfway, or looks square, is not a median or an altitude until tick marks, a square mark or the words say so.
There must be a triangle. If A, B and C lie on one line, the “altitudes” have length zero and the area is zero. Every fact here assumes the three points are not on one line.
Where this goes next
The three medians of any triangle pass through a single point, and that point sits two thirds of the way down each median from its vertex. It is the balance point of a cardboard triangle, the Centroid. The three altitude lines also pass through a single point, the orthocenter, which wanders outside the triangle in exactly the obtuse case seen here.
The altitude to the hypotenuse of a right triangle cuts it into two smaller triangles that are the same shape as the whole, and that gives the geometric mean relations in a right triangle.
A median joins a vertex to a midpoint. Joining two midpoints instead gives a midsegment, which is parallel to the third side and half as long. And the isosceles result in this lesson is the starting point of the proof that base angles are equal, which adds exactly this median to split the triangle into two congruent halves. The altitude itself is the triangle’s case of a general idea, the distance from a point to a line.
Exercises
Naming the segments
- In \triangle PQR, S is the midpoint of \overline{QR}, and T is the point of \overline{QR} with \overline{PT} \perp \overline{QR}. Name the median from P and the altitude from P.
- True or false: every altitude of a triangle lies inside the triangle. Explain.
- True or false: a median is always perpendicular to the side it meets. Explain.
Medians
- In \triangle ABC, \overline{AD} is a median, BD = 2x + 7 and DC = 4x - 9. Find x and BC.
- In \triangle JKL, \overline{JM} is a median, KM = 3y - 1 and KL = 4y + 8. Find y, KM and KL.
- \overline{AM} is a median of \triangle ABC. The area of \triangle ABM is 3k + 5 and the area of \triangle ABC is 8k - 10. Find k and both areas.
- The triangle has vertices P(-3, 2), Q(5, 8) and R(1, -4). Find the lengths of the medians from P, from R and from Q.
- In \triangle ABC, A(2, 9) and B(-4, 1), and the median from A ends at M(1, 3). Find the coordinates of C.
Altitudes
- In \triangle ABC, \overline{AD} is an altitude with D on \overline{BC}, m\angle ADC = (4x + 10)^\circ and m\angle DAC = (2x + 3)^\circ. Find x, m\angle DAC and m\angle C.
- A triangle has a side of length 14 whose altitude is 9. Another side has length 18. Find the area and the altitude to the side of length 18.
- A triangle has sides 15, 20 and 25. Find the lengths of all three altitudes.
- The triangle has vertices A(-2, 7), B(-5, 1) and C(4, 1). Find the foot and length of the altitude from A, and the area of the triangle.
- The triangle has vertices A(-4, 3), B(0, 0) and C(5, 0). Find the foot of the altitude from A, say whether it lies on \overline{BC}, and find the area and the length of the altitude from C.
- In \triangle ABC, AB = 15, AC = 13 and BC = 14. The altitude from A has its foot D on \overline{BC}. Let BD = x. Use the The Pythagorean theorem in both right triangles to find x, then find AD and the area.
Isosceles and right triangles
- In \triangle ABC, AB = AC = 10 and BC = 12. Find the length of the median from A, the area, and the altitude from B.
- In \triangle ABC, AB = AC and \overline{AM} is the median to \overline{BC}. If m\angle BAM = 34^\circ, find m\angle CAM, m\angle AMB and m\angle B.
- A right triangle has legs 9 and 12. Find the median to the hypotenuse and the altitude to the hypotenuse.
- In a right triangle the median to the hypotenuse is 8.5 and one leg is 8. Find the other leg.
- \triangle ABC has area 60. \overline{AM} is a median and N is the midpoint of \overline{AM}. Find the area of \triangle BNC.
Reasoning
- Explain why a median can never lie outside its triangle, while an altitude can.
- In \triangle ABC the altitude from A also bisects \angle BAC. Prove that AB = AC.
- Explain why the altitude from A is never longer than \overline{AB}, and say when the two are equally long.
- Can a triangle have exactly one altitude that lies outside it? Explain.
- Show that in an equilateral triangle each median is also an altitude.
- The three altitudes of a triangle have lengths 3, 4 and 6. Which altitude goes to the longest side? Find the ratio of the three sides, longest first.
Answers
- The median is \overline{PS}, because S is the midpoint of the opposite side. The altitude is \overline{PT}, because it is perpendicular to the opposite side.
- False. In an obtuse triangle the altitudes from the two acute vertices land on sides extended past the obtuse corner, so they lie outside. (In acute and right triangles none lie outside.)
- False. A median lands at the midpoint, which says nothing about the angle. Example 2 has a median that is not perpendicular to its side; median and altitude coincide only when the two sides from the vertex are equal.
- x = 8 and BC = 46. D is the midpoint, so 2x + 7 = 4x - 9, giving 2x = 16. Then BD = DC = 23.
- y = 5, KM = 14 and KL = 28. M is the midpoint, so KL = 2 \cdot KM: 4y + 8 = 2(3y - 1) = 6y - 2, so 2y = 10.
- k = 10, with areas 35 and 70. The median halves the area, so 8k - 10 = 2(3k + 5), which gives 8k - 10 = 6k + 10 and 2k = 20.
- 6, 9 and \displaystyle 3\sqrt{13}. The midpoint of \overline{QR} is (3, 2), which is 6 from P(-3, 2). The midpoint of \overline{PQ} is (1, 5), which is 9 from R(1, -4). The midpoint of \overline{PR} is (-1, -1), and its distance from Q(5, 8) is \displaystyle \sqrt{6^2 + 9^2} = \sqrt{117} = 3\sqrt{13}.
- C(6, 5). M is the midpoint of \overline{BC}, so each coordinate of M is the average of those of B and C: C = (2 \cdot 1 - (-4),\ 2 \cdot 3 - 1).
- x = 20, m\angle DAC = 43^\circ and m\angle C = 47^\circ. An altitude makes a right angle, so 4x + 10 = 90. Then 2(20) + 3 = 43, and the angles of \triangle ADC give 180 - 90 - 43 = 47.
- Area 63; the altitude to the side of length 18 is 7. The area is \tfrac12 \cdot 14 \cdot 9, and \tfrac12 \cdot 18 \cdot h = 63 gives \displaystyle h = \frac{126}{18}.
- 20, 15 and 12. Since 15^2 + 20^2 = 625 = 25^2, it is a right triangle, so each leg is the altitude to the other: the altitude to the side 15 is 20, and to the side 20 is 15. The area is \tfrac12 \cdot 15 \cdot 20 = 150, so the altitude to the hypotenuse is \displaystyle \frac{300}{25} = 12.
- Foot (-2, 1), length 6, area 27. \overline{BC} is horizontal, so the altitude is vertical; the foot is between x = -5 and x = 4, the length is 7 - 1, and BC = 9, so the area is \tfrac12 \cdot 9 \cdot 6.
- The foot is (-4, 0), which is not on \overline{BC}: it is 4 units beyond B. The area is \tfrac12 \cdot 5 \cdot 3 = 7.5. With \displaystyle AB = \sqrt{4^2 + 3^2} = 5 as the base, the altitude from C is \displaystyle \frac{2 \cdot 7.5}{5} = 3. (This triangle has BA = BC = 5, so the altitudes from A and C are equally long, as they should be.)
- x = 9, AD = 12 and the area is 84. Both right triangles share AD, so 15^2 - x^2 = 13^2 - (14 - x)^2. Expanding, 225 - x^2 = -27 + 28x - x^2, so 28x = 252. Then \displaystyle AD = \sqrt{225 - 81} = 12 and the area is \tfrac12 \cdot 14 \cdot 12.
- Median 8, area 48, altitude from B equal to 9.6. BM = 6 and the median is perpendicular to \overline{BC}, so it is \displaystyle \sqrt{10^2 - 6^2} = 8. The area is \tfrac12 \cdot 12 \cdot 8, and \displaystyle h_B = \frac{96}{10}.
- m\angle CAM = 34^\circ, m\angle AMB = 90^\circ and m\angle B = 56^\circ. In an isosceles triangle the median from the apex bisects the apex angle and is perpendicular to the base; then 180 - 90 - 34 = 56.
- Median 7.5, altitude 7.2. The hypotenuse is \displaystyle \sqrt{81 + 144} = 15, and the median to it is half of that. The area is \tfrac12 \cdot 9 \cdot 12 = 54, so the altitude is \displaystyle \frac{108}{15}.
- 15. The hypotenuse is twice the median, 17, so the other leg is \displaystyle \sqrt{17^2 - 8^2} = \sqrt{225}.
- 30. \triangle ABM and \triangle AMC each have area 30. In \triangle ABM, \overline{BN} is a median, so \triangle BNM has area 15; in the same way \triangle CNM has area 15. These two make up \triangle BNC.
- A median ends at the midpoint of a side, which is always a point of the triangle, so the segment from the vertex to it runs through the inside. An altitude ends wherever the perpendicular meets the line of the opposite side, and when the angle at an end of that side is obtuse, that point is beyond the side, outside the triangle.
- Let D be the foot. The altitude lies along the bisector of \angle BAC, which runs through the inside of the triangle, so D is strictly between B and C and both triangles below exist. In \triangle ABD and \triangle ACD: \angle BAD \cong \angle CAD because \overline{AD} bisects \angle BAC; AD = AD; and \angle ADB \cong \angle ADC because both are right angles. By ASA the triangles are congruent, so AB = AC as corresponding parts.
- If the foot D is not B, then \triangle ABD has a right angle at D, so \overline{AB} is its hypotenuse and AB^2 = AD^2 + BD^2 > AD^2, so AD < AB. If D = B, the altitude is \overline{AB} itself. So the two are equal exactly when \angle B is a right angle.
- No. An altitude lies outside exactly when one end of its side has an obtuse angle. A triangle has at most one obtuse angle. If it has none, no altitude lies outside. If \angle B is obtuse, the altitudes from A and from C both lie outside (each side through B has an obtuse end at B), and the altitude from B lies inside, because \angle A and \angle C are acute. So the count is 0 or 2.
- In an equilateral triangle AB = AC, so the median from A is an altitude by the isosceles result. Also BA = BC, so the median from B is an altitude, and CA = CB does the same for C.
- The altitude 3 goes to the longest side, and the sides are in the ratio 4 : 3 : 2. Each side times its altitude is twice the area, so each side is \displaystyle \frac{2A}{h}: the sides are \displaystyle \frac{2A}{3}, \displaystyle \frac{2A}{4} and \displaystyle \frac{2A}{6}. The smallest altitude gives the largest side, and multiplying through by \displaystyle \frac{6}{A} gives 4 : 3 : 2.