Geometry · G.C.A.2

Angles made by chords, secants and tangents

An angle at the center of a circle equals its arc, and an angle on the circle is half its arc. What happens when the vertex moves inside the circle, or outside it, or when one side is a tangent?

You need

The inscribed angle theoremGeometry Tangent propertiesGeometry Vertical anglesMath 7

Where you are 15 in Circles

Geometry11

Circles15

▴ 8 above Sectors, segments and their areas The inscribed angle theorem

Angles made by chords, secants and tangents

Chord theorems Segment lengths in circles ▾ Show all 15 (2 more)

Why this exists

A central angle has its vertex at the center of a circle, and it measures the same as its arc. An inscribed angle has its vertex on the circle, and it measures half its arc. Those are only two places for a vertex. Two chords can cross anywhere inside a circle, two secants can meet anywhere outside it, and a side of an angle can be a tangent that touches the circle instead of crossing it.

Every one of these angles cuts off arcs, and the arcs decide the angle. This lesson finds the rule for each position. A vertex inside the circle gives half the sum of two arcs, a vertex outside gives half their difference, and a tangent behaves like a chord whose two ends have run together. Together with central and inscribed angles, that covers every angle whose two sides cross or touch a circle.

Definition

Recall from chords, secants and tangents that a secant is a line that meets a circle at two points, and a tangent is a line that meets it at exactly one point, its point of contact. The angles in this lesson have their sides along chords, secants or tangents, and they are sorted by where the vertex sits.

A tangent-chord angle has its vertex at the point of contact A of a tangent, one side along the tangent and the other along a chord \overline{AB}. Its intercepted arc is the arc from A to B that lies inside the angle. In Figure 1, \angle QAB intercepts the heavy minor arc \overset{\frown}{AB}, and the angle on the other side of the chord, \angle PAB, intercepts the major arc.

Figure 1The tangent-chord angle \angle QAB intercepts the heavy arc \overset{\frown}{AB}. The dashed radii are used in the proof.

An angle formed by two chords has its vertex at the point E where two chords \overline{AC} and \overline{BD} cross inside the circle. The angle \angle AEB intercepts the arc \overset{\frown}{AB} that lies inside it, and its vertical angle \angle CED intercepts \overset{\frown}{CD}. The two angles are equal, so both arcs belong to the one size of angle at E: the arc in front of the vertex and the arc behind it.

An angle with its vertex outside the circle has its vertex at a point P outside, and each of its sides is a secant or a tangent. That gives three kinds, shown in Figure 2: two secants, a secant and a tangent, and two tangents. Between its sides the angle cuts off two arcs: the far arc, on the side of the circle away from P, and the near arc, facing P. A point of contact is an end of both arcs. With two tangents the two arcs make up the whole circle, the near arc being the minor arc and the far arc the major arc. That angle is the circumscribed angle of tangent properties.

Figure 2Three kinds of angle with the vertex outside. Each cuts off a far arc and a near arc.

Arcs here are measured in degrees, as always: the measure of an arc is the measure of its central angle.

Intuition

Fix two points A and B on a circle. Put a vertex V on the line through the center and the middle of \overset{\frown}{AB}, keep its two sides through A and B, and slide it along that line, as in Figure 3.

Figure 3One arc seen from inside the circle, from on it and from outside it. Inside, the angle gains the arc behind; outside, it loses the arc in front.

With V on the circle, in panel (b), \angle AVB is an inscribed angle: half of \overset{\frown}{AB}.

Push V inside, as in panel (a). The angle opens wider. Its sides, carried on through V, now cut off a second arc \overset{\frown}{CD} behind the vertex, and the angle collects half of each arc: half the sum.

Pull V outside, as in panel (c). The angle closes up. Its sides cross the circle at C and D before they reach A and B, cutting off a near arc between V and the far arc, and that near arc counts against the angle: half the far arc minus half the near arc.

The three rules join up with no jump. As V comes onto the circle from inside, the arc behind it shrinks to nothing, and half the sum becomes half of \overset{\frown}{AB}. As V comes onto the circle from outside, the near arc shrinks to nothing, and half the difference becomes half of \overset{\frown}{AB} as well. At the center, the arc in front and the arc behind are equal, so half their sum is the arc itself, the central angle.

A tangent fits the same picture. Turn the chord \overline{AB} about A until B runs into A: the line through them becomes the tangent at A, and the arc between A and B shrinks to nothing. So a tangent side acts like a chord whose ends have merged, and each rule carries over to it with that arc taken as 0^\circ.

Results

Tangent-chord angle. The angle between a tangent and a chord drawn from the point of contact is half its intercepted arc. In Figure 1,

m\angle QAB = \tfrac{1}{2}\, m\overset{\frown}{AB}

Two chords crossing inside. If chords \overline{AC} and \overline{BD} cross at E, then

m\angle AEB = \tfrac{1}{2}\left(m\overset{\frown}{AB} + m\overset{\frown}{CD}\right)

Vertex outside. If two secants, a secant and a tangent, or two tangents meet at a point P outside a circle, then

m\angle P = \tfrac{1}{2}\left(\text{far arc} - \text{near arc}\right)

For two tangents whose near arc, the minor arc, measures m, the far arc is 360^\circ - m, and the rule becomes the form found in tangent properties:

m\angle P = \tfrac{1}{2}\big((360^\circ - m) - m\big) = 180^\circ - m

The whole family, sorted by where the vertex is:

Where the vertex is	The angle
at the center	the arc
inside the circle	half the sum of the arc in front and the arc behind
on the circle (inscribed, or tangent and chord)	half the arc
outside the circle	half the far arc minus half the near arc

Why it is true

Every case rests on two facts already proved: the inscribed angle theorem, and the exterior angle theorem, which says that an angle outside a triangle, next to one of its angles along a straight side, equals the sum of the two angles of the triangle that are not next to it. The tangent-chord case comes first, because the outside cases with a tangent use it.

Tangent and chord. In Figure 1, let the minor arc \overset{\frown}{AB} measure x, so the central angle \angle AOB measures x as well, and let y = m\angle QAB. Triangle OAB has two radii for sides, so it is isosceles and its base angles are equal. They share what is left of 180^\circ:

m\angle OAB = \tfrac{1}{2}\left(180^\circ - x\right) = 90^\circ - \tfrac{x}{2}

A tangent is perpendicular to the radius at its point of contact, so m\angle QAO = 90^\circ. The chord runs inside that right angle, because B is on the same side of line OA as Q, so by angle addition

y = 90^\circ - \left(90^\circ - \tfrac{x}{2}\right) = \tfrac{x}{2}

The angle on the other side of the chord, \angle PAB, forms a linear pair with \angle QAB, so it measures 180^\circ - \tfrac{x}{2} = \tfrac{1}{2}\left(360^\circ - x\right), which is half the major arc it intercepts. If \overline{AB} is a diameter, both angles are 90^\circ, half of a semicircle. So the rule holds in every case.

Two chords. In Figure 4(a), draw the chord \overline{BC}. The angle x = m\angle AEB sits next to the angle of triangle EBC at E, along the straight line AC, so it is an exterior angle of that triangle and

x = y + z

where y = m\angle ACB and z = m\angle DBC are the two angles of the triangle not next to it. Both are inscribed: \angle ACB intercepts \overset{\frown}{AB} and \angle DBC intercepts \overset{\frown}{CD}. So y = \tfrac12\, m\overset{\frown}{AB} and z = \tfrac12\, m\overset{\frown}{CD}, and adding them,

m\angle AEB = \tfrac{1}{2}\left(m\overset{\frown}{AB} + m\overset{\frown}{CD}\right)

Figure 4Each proof draws one chord and uses an exterior angle of a triangle. In (a) x = y + z; in (b) z = x + y.

Two secants. In Figure 4(b), the secants from P meet the circle at A and C and at B and D, with A and B nearer to P. Draw the chord \overline{AD}. The points P, A and C lie on one line, so z = m\angle CAD is an exterior angle of triangle PAD at A, and

z = x + y

where x = m\angle P and y = m\angle ADP, which is the same angle as \angle ADB because B lies on \overline{DP}. Both z and y are inscribed angles: \angle CAD intercepts the far arc \overset{\frown}{CD}, and \angle ADB intercepts the near arc \overset{\frown}{AB}. Therefore

m\angle P = z - y = \tfrac{1}{2}\, m\overset{\frown}{CD} - \tfrac{1}{2}\, m\overset{\frown}{AB}

A secant and a tangent. Let the tangent touch at T and the secant meet the circle at A and then C. Draw the chord \overline{TA}. As before, \angle TAC is an exterior angle of triangle PTA, so m\angle TAC = m\angle P + m\angle PTA. Here \angle TAC is inscribed and intercepts the far arc \overset{\frown}{TC}, while \angle PTA is a tangent-chord angle that intercepts the near arc \overset{\frown}{TA}. So m\angle P = \tfrac12\, m\overset{\frown}{TC} - \tfrac12\, m\overset{\frown}{TA}.

Two tangents. Let them touch at S and T. Draw the chord \overline{ST}, and extend \overline{PS} beyond S to a point Q. Then \angle QST is an exterior angle of triangle PST, so m\angle QST = m\angle P + m\angle PTS. Both are tangent-chord angles: \angle QST opens away from P and intercepts the far arc, and \angle PTS opens toward P and intercepts the near arc. So once more m\angle P is half the far arc minus half the near arc.

The tangent-chord fact is old. Euclid proves it as Proposition 32 of Book III of the Elements, in the form used in Exercise 15 below: the angle between a tangent and a chord equals any inscribed angle standing on the other side of the chord.

Worked examples

Example 1

A tangent and a chord

A tangent touches a circle at A, and \overline{AB} is a chord. One of the two angles the chord makes with the tangent intercepts an arc of 128^\circ. Find both angles.

Solution

A tangent-chord angle is half its intercepted arc:

\tfrac{1}{2} \cdot 128^\circ = 64^\circ

The other angle, on the far side of the chord, intercepts the rest of the circle, 360^\circ - 128^\circ = 232^\circ, so it measures \tfrac{1}{2} \cdot 232^\circ = 116^\circ.

Check. The two angles sit side by side along the tangent, a linear pair, and 64^\circ + 116^\circ = 180^\circ.

Example 2

Two chords crossing

In Figure 5, chords \overline{AC} and \overline{BD} cross at E, with m\overset{\frown}{AB} = 86^\circ and m\overset{\frown}{CD} = 120^\circ. Find m\angle AEB, m\angle CED and m\angle BEC.

Figure 5Example 2. Chords \overline{AC} and \overline{BD} cross at E.

Solution

The vertex is inside the circle, so the angle is half the sum of the arc in front and the arc behind. \angle AEB intercepts \overset{\frown}{AB}, and its vertical angle intercepts \overset{\frown}{CD}:

m\angle AEB = \tfrac{1}{2}\left(86^\circ + 120^\circ\right) = \tfrac{1}{2} \cdot 206^\circ = 103^\circ

\angle CED is vertical to \angle AEB, so m\angle CED = 103^\circ. \angle BEC forms a linear pair with \angle AEB, so m\angle BEC = 180^\circ - 103^\circ = 77^\circ.

Check. \angle BEC is itself an angle formed by the two chords. It intercepts \overset{\frown}{BC}, and its vertical angle intercepts \overset{\frown}{DA}. Those two arcs are what is left of the circle, 360^\circ - 86^\circ - 120^\circ = 154^\circ, and half of 154^\circ is 77^\circ, the same answer by a second road.

Example 3

An arc from the angle

Chords \overline{AC} and \overline{BD} cross at E, and m\angle AEB = 75^\circ. The arcs are m\overset{\frown}{AB} = (2x + 8)^\circ and m\overset{\frown}{CD} = (4x - 2)^\circ. Find x and both arcs.

Solution

The angle is half the sum of the two arcs, so

\begin{aligned} \tfrac{1}{2}\big((2x + 8) + (4x - 2)\big) &= 75 \\ 6x + 6 &= 150 \\ x &= 24 \end{aligned}

Then m\overset{\frown}{AB} = 2(24) + 8 = 56^\circ and m\overset{\frown}{CD} = 4(24) - 2 = 94^\circ.

Check. \tfrac12(56 + 94) = \tfrac12 \cdot 150 = 75.

Example 4

Two secants from outside

In Figure 6, two secants from P cross the circle at A and C and at B and D. The far arc \overset{\frown}{CD} measures 150^\circ and the near arc \overset{\frown}{AB} measures 40^\circ. Find m\angle P.

Figure 6Example 4. Two secants from P, with a far arc of 150^\circ and a near arc of 40^\circ.

Solution

The vertex is outside, so the angle is half the far arc minus half the near arc:

m\angle P = \tfrac{1}{2}\left(150^\circ - 40^\circ\right) = \tfrac{1}{2} \cdot 110^\circ = 55^\circ

Check. Run the proof with these numbers. The inscribed angle \angle CAD on the far arc is \tfrac12 \cdot 150^\circ = 75^\circ, and the inscribed angle \angle ADB on the near arc is \tfrac12 \cdot 40^\circ = 20^\circ. The exterior angle at A is the sum of the other two angles of triangle PAD: 75^\circ = 55^\circ + 20^\circ. From A on the circle the arc \overset{\frown}{CD} is seen at 75^\circ; from P, farther off, it is seen at only 55^\circ.

Example 5

A secant and a tangent

From a point P, a tangent touches a circle at T, and a secant meets the circle first at A and then at C. m\angle P = 32^\circ and the near arc \overset{\frown}{TA} measures 64^\circ. Find the far arc \overset{\frown}{TC} and the third arc \overset{\frown}{AC}, the one that does not contain T.

Solution

The vertex is outside, so

\begin{aligned} \tfrac{1}{2}\left(m\overset{\frown}{TC} - 64\right) &= 32 \\ m\overset{\frown}{TC} - 64 &= 64 \\ m\overset{\frown}{TC} &= 128 \end{aligned}

The far arc measures 128^\circ. The points T, A and C cut the circle into three arcs, which fill it, so

m\overset{\frown}{AC} = 360^\circ - 64^\circ - 128^\circ = 168^\circ

Example 6

Two tangents

Two tangents from P touch a circle at S and T. The major arc and the minor arc from S to T are in the ratio 7 : 5. Find m\angle P and m\angle PST.

Solution

Write the arcs as 7k and 5k. Together they make the whole circle:

12k = 360 \qquad\Longrightarrow\qquad k = 30

so the major arc is 210^\circ and the minor arc 150^\circ. The far arc is the major one and the near arc the minor one:

m\angle P = \tfrac{1}{2}\left(210^\circ - 150^\circ\right) = 30^\circ

which agrees with 180^\circ - 150^\circ = 30^\circ.

\angle PST is a tangent-chord angle. It opens toward P, so it intercepts the near arc:

m\angle PST = \tfrac{1}{2} \cdot 150^\circ = 75^\circ

Check. The two tangent segments from one point are equal, PS = PT, so triangle PST is isosceles with apex angle 30^\circ, and each base angle is \tfrac12\left(180^\circ - 30^\circ\right) = 75^\circ.

Example 7

Four points, both kinds of angle

As in Figure 6, two secants from P cross a circle at A and C and at B and D, with A and B nearer to P. Now m\angle P = 25^\circ, the near arc is m\overset{\frown}{AB} = x^\circ, the far arc is m\overset{\frown}{CD} = (3x - 10)^\circ, and the two side arcs \overset{\frown}{AC} and \overset{\frown}{BD} are equal. Find all four arcs, and the angles at the point E where the chords \overline{AD} and \overline{BC} cross.

Solution

The vertex P is outside:

\begin{aligned} \tfrac{1}{2}\big((3x - 10) - x\big) &= 25 \\ 2x - 10 &= 50 \\ x &= 30 \end{aligned}

So m\overset{\frown}{AB} = 30^\circ and m\overset{\frown}{CD} = 3(30) - 10 = 80^\circ. The four arcs fill the circle, and the two side arcs are equal, so each is

\tfrac{1}{2}\left(360^\circ - 30^\circ - 80^\circ\right) = 125^\circ

Now the chords. \overline{AD} and \overline{BC} cross inside the circle at E. The angle \angle AEB opens onto \overset{\frown}{AB}, and its vertical angle \angle DEC onto \overset{\frown}{CD}, so

m\angle AEB = \tfrac{1}{2}\left(30^\circ + 80^\circ\right) = 55^\circ

The neighboring angle \angle AEC opens onto \overset{\frown}{AC}, and its vertical angle onto \overset{\frown}{BD}:

m\angle AEC = \tfrac{1}{2}\left(125^\circ + 125^\circ\right) = 125^\circ

Check. 55^\circ + 125^\circ = 180^\circ, as a linear pair must. Notice also that m\angle AEB - m\angle P = 55^\circ - 25^\circ = 30^\circ, the near arc. That is always so for this arrangement: half the sum minus half the difference of two numbers is the smaller one.

Example 8

A tangent parallel to a chord

Triangle ABC is inscribed in a circle, and the tangent at A is parallel to \overline{BC}. Show that AB = AC.

Solution

Choose a point Q on the tangent so that Q and C lie on opposite sides of line AB. The line AB crosses the two parallel lines, the tangent and \overleftrightarrow{BC}, and \angle QAB and \angle ABC are alternate interior angles, so they are equal.

\angle QAB is a tangent-chord angle. The arc inside it lies on Q’s side of line AB, so it is the arc \overset{\frown}{AB} that does not contain C, and m\angle QAB = \tfrac12\, m\overset{\frown}{AB}. The inscribed angle \angle ACB intercepts that same arc, so m\angle ACB = \tfrac12\, m\overset{\frown}{AB} as well.

Putting the two facts together,

m\angle ABC = m\angle QAB = m\angle ACB

Triangle ABC has two equal angles, at B and at C, so by the converse of the base angles theorem the sides opposite them are equal: AC = AB. In other words, A sits at the middle of the arc \overset{\frown}{BC} on its side.

Boundaries

Which rule depends only on where the vertex is. Find the vertex before reaching for a formula. On the circle the angle uses one arc; inside it uses the arc in front and the arc behind, added; outside it uses the far arc and the near arc, subtracted.

Only the arcs between the sides count. For a vertex outside, the far and near arcs are the ones lying between the two sides. The other arcs, such as \overset{\frown}{AC} and \overset{\frown}{BD} in Example 7, do not enter the rule for \angle P. For a vertex inside, the two arcs are the one inside the angle and the one inside its vertical angle, not the two beside them.

Each side must meet the circle. If one side of an angle at an outside point misses the circle, it cuts off no arc, and the angle can be anything as that side swings. None of these rules applies.

A tangent-chord angle has its vertex at the point of contact. If a tangent at T meets a chord that does not end at T, the chord’s line crosses the tangent somewhere else, at a point outside the circle, and the angle there is a secant-tangent angle, found with the difference rule.

Parallel lines make no angle. Two parallel secants never meet, so there is no vertex. The difference rule still points the right way: the two arcs between parallel chords are equal, as inscribed angles show, which is what a difference of 0^\circ would give.

Degrees of arc, not length. Every arc here is measured in degrees, by its central angle. The rules say nothing about arc lengths, which depend on the size of the circle.

Where this goes next

The figures in this lesson also carry lengths. When two chords cross, the inscribed angles on equal arcs make the two small triangles at the crossing similar, and the same happens for two secants or a secant and a tangent from an outside point. The similar triangles turn into the product rules of segment lengths in circles, drawn on exactly these pictures.

The angle rules also feed straight back into the rest of circle geometry. Opposite angles of a quadrilateral inscribed in a circle, the angle between its diagonals, and the angles a tangent makes with a triangle inscribed in the circle are all found this way. The method of the proofs, one extra chord and an exterior angle, is a pattern used again and again in proofs about circles.

Exercises

A tangent and a chord

A tangent and a chord meet at the point of contact, and the angle between them intercepts an arc of 76^\circ. Find the angle.
The angle between a tangent and a chord at the point of contact is 67^\circ. Find its intercepted arc, and the arc on the other side of the chord.
The angle between a tangent at A and a chord \overline{AB} is 112^\circ. Find the measure of the minor arc \overset{\frown}{AB}.

Two chords crossing

In Exercises 4 to 7, chords \overline{AC} and \overline{BD} cross at E, and A, B, C, D lie on the circle in that order.

m\overset{\frown}{AB} = 54^\circ and m\overset{\frown}{CD} = 98^\circ. Find m\angle AEB and m\angle BEC.
m\angle AEB = 64^\circ and m\overset{\frown}{AB} = 50^\circ. Find m\overset{\frown}{CD}.
Going round the circle, the arcs \overset{\frown}{AB}, \overset{\frown}{BC}, \overset{\frown}{CD} and \overset{\frown}{DA} measure 72^\circ, 110^\circ, 96^\circ and 82^\circ. Find all four angles at E.
m\overset{\frown}{AB} = (3x)^\circ, m\overset{\frown}{CD} = (x + 40)^\circ and m\angle AEB = 70^\circ. Find x and both arcs.

The vertex outside

Two secants from P cut off a far arc of 130^\circ and a near arc of 46^\circ. Find m\angle P.
A secant and a tangent from P cut off a far arc of 200^\circ and a near arc of 70^\circ. Find m\angle P.
Two tangents from P touch a circle, and the minor arc between the points of contact measures 110^\circ. Find the major arc and m\angle P.
Two secants from P make an angle of 30^\circ, and the far arc measures 110^\circ. Find the near arc.
Two tangents from P make an angle of (x + 20)^\circ, and the minor arc between the points of contact measures (3x)^\circ. Find x, m\angle P and the minor arc.
From P, a tangent touches a circle at T and a secant meets it at A and then C. The near arc \overset{\frown}{TA} measures 50^\circ, and the arc \overset{\frown}{AC} that does not contain T measures 150^\circ. Find the far arc \overset{\frown}{TC} and m\angle P.

More than one step

A, B, C, D lie on a circle in that order, with m\overset{\frown}{AB} = 50^\circ, m\overset{\frown}{BC} = 130^\circ, m\overset{\frown}{CD} = 70^\circ and m\overset{\frown}{DA} = 110^\circ. The chords \overline{AC} and \overline{BD} cross at E, and the lines BA and CD meet at a point P beyond A and D. Find m\angle AEB, m\angle BEC and m\angle P.
Triangle ABC is inscribed in a circle with m\overset{\frown}{AB} = 100^\circ, m\overset{\frown}{BC} = 140^\circ and m\overset{\frown}{CA} = 120^\circ, each arc taken as the one not containing the third vertex. A tangent is drawn at A. Find the angle between the tangent and \overline{AB} on the side away from C, and compare it with m\angle ACB.
Two tangents from P touch a circle at S and T. A point Q on the major arc makes m\angle SQT = 65^\circ. Find m\angle P.
From P, a tangent touches a circle with center O at T, and a secant through O meets the circle at A and then C. m\angle P = 40^\circ. Find m\overset{\frown}{TA} and m\overset{\frown}{TC}, the arcs from T to each end of the diameter \overline{AC}.

Reasoning

Chords \overline{AC} and \overline{BD} cross at the center of the circle. Show that the rule for two chords gives the central angle, as it must.
A vertex P outside a circle moves toward the circle, with its sides always passing through two fixed points C and D on the far arc. Explain what happens to the near arc, and why the outside rule becomes the inscribed angle theorem when P reaches the circle.
Two tangents from P cut off a minor arc of m degrees. Show that the outside rule gives m\angle P = 180^\circ - m, the circumscribed angle found from the quadrilateral OAPB in tangent properties.
Two secants from P cross a circle at A and C and at B and D, with A and B nearer to P. Show that m\angle P < m\angle CAD, and say in words what this means about seeing an arc from on the circle and from outside it.
Chords \overline{AC} and \overline{BD} cross at right angles. Show that m\overset{\frown}{AB} + m\overset{\frown}{CD} = 180^\circ, and that the other two arcs also add to 180^\circ.
Two tangents from P touch a circle at S and T, and m\angle P = 60^\circ. Show that triangle PST is equilateral.

Answers

38^\circ. A tangent-chord angle is half its arc: \tfrac12 \cdot 76^\circ = 38^\circ.
The intercepted arc is 134^\circ and the other arc is 226^\circ. The arc is twice the angle, 2 \cdot 67^\circ = 134^\circ, and the rest of the circle is 360^\circ - 134^\circ = 226^\circ.
136^\circ. The angle is obtuse, so it intercepts the major arc, which measures 2 \cdot 112^\circ = 224^\circ. The minor arc is 360^\circ - 224^\circ = 136^\circ. Check: the angle on the other side of the chord is 180^\circ - 112^\circ = 68^\circ, half of 136^\circ.
m\angle AEB = 76^\circ and m\angle BEC = 104^\circ. Half the sum is \tfrac12(54^\circ + 98^\circ) = 76^\circ, and \angle BEC forms a linear pair with \angle AEB, so it is 180^\circ - 76^\circ = 104^\circ.
78^\circ. From 64 = \tfrac12\left(50 + m\overset{\frown}{CD}\right), 128 = 50 + m\overset{\frown}{CD}.
m\angle AEB = m\angle CED = 84^\circ and m\angle BEC = m\angle DEA = 96^\circ. Each is half the sum of its arc and the arc of its vertical angle: \tfrac12(72 + 96) = 84 and \tfrac12(110 + 82) = 96. Check: 84 + 96 = 180, as a linear pair must.
x = 25, m\overset{\frown}{AB} = 75^\circ and m\overset{\frown}{CD} = 65^\circ. From \tfrac12\big(3x + (x + 40)\big) = 70, 4x + 40 = 140.
42^\circ. \tfrac12(130^\circ - 46^\circ) = \tfrac12 \cdot 84^\circ = 42^\circ.
65^\circ. \tfrac12(200^\circ - 70^\circ) = \tfrac12 \cdot 130^\circ = 65^\circ.
The major arc is 250^\circ and m\angle P = 70^\circ. The two arcs fill the circle, so the major arc is 360^\circ - 110^\circ, and \tfrac12(250^\circ - 110^\circ) = 70^\circ, the same as 180^\circ - 110^\circ.
50^\circ. From 30 = \tfrac12(110 - n), 60 = 110 - n, so the near arc is n = 50.
x = 40, m\angle P = 60^\circ and the minor arc is 120^\circ. For two tangents m\angle P = 180^\circ minus the minor arc, so x + 20 = 180 - 3x and 4x = 160. Check: \tfrac12(240 - 120) = 60.
m\overset{\frown}{TC} = 160^\circ and m\angle P = 55^\circ. The arcs \overset{\frown}{TA}, \overset{\frown}{AC} and \overset{\frown}{CT} fill the circle, so m\overset{\frown}{TC} = 360^\circ - 50^\circ - 150^\circ = 160^\circ, and \tfrac12(160^\circ - 50^\circ) = 55^\circ.
m\angle AEB = 60^\circ, m\angle BEC = 120^\circ and m\angle P = 10^\circ. Inside: \tfrac12(50 + 70) = 60 and \tfrac12(130 + 110) = 120. Outside: between the lines BA and CD lie the arcs \overset{\frown}{BC} and \overset{\frown}{DA}, and the near one is \overset{\frown}{DA}, since P is beyond A and D, so \tfrac12(130 - 110) = 10.
Both are 50^\circ. The tangent-chord angle on the side away from C intercepts \overset{\frown}{AB}, so it is \tfrac12 \cdot 100^\circ = 50^\circ. \angle ACB is inscribed and intercepts the same arc, so it is also 50^\circ. This is Euclid’s form of the theorem.
50^\circ. Q is on the major arc, so \angle SQT intercepts the minor arc, which measures 2 \cdot 65^\circ = 130^\circ. Then m\angle P = 180^\circ - 130^\circ = 50^\circ.
m\overset{\frown}{TA} = 50^\circ and m\overset{\frown}{TC} = 130^\circ. The two arcs together make the semicircle from A through T to C, so they add to 180^\circ. The outside rule says their difference is 2 \cdot 40^\circ = 80^\circ. Adding the two equations, 2\, m\overset{\frown}{TC} = 260^\circ. Check: \overline{OT} \perp \overline{PT}, so in the right triangle OTP the angle at O is 90^\circ - 40^\circ = 50^\circ, the central angle on \overset{\frown}{TA}.
With E at the center, \angle AEB and \angle CED are central angles, so they measure m\overset{\frown}{AB} and m\overset{\frown}{CD}. They are vertical angles, so they are equal, and the two arcs are equal. Then \tfrac12\left(m\overset{\frown}{AB} + m\overset{\frown}{CD}\right) = \tfrac12 \cdot 2\, m\overset{\frown}{AB} = m\overset{\frown}{AB}, which is the central angle.
As P comes closer, the near points where the sides first cross the circle slide toward each other and toward the spot where P will land, so the near arc shrinks. When P reaches the circle, the near arc has shrunk to nothing, 0^\circ, and the sides are chords \overline{PC} and \overline{PD}. The outside rule then reads \tfrac12\left(m\overset{\frown}{CD} - 0\right) = \tfrac12\, m\overset{\frown}{CD}, which is the inscribed angle theorem.
The far arc is the rest of the circle, 360 - m. So \tfrac12\big((360 - m) - m\big) = \tfrac12(360 - 2m) = 180 - m.
\angle CAD is an exterior angle of triangle PAD, so m\angle CAD = m\angle P + m\angle ADP. The angle \angle ADP is an angle of a triangle, so it is more than 0^\circ, and m\angle CAD is larger than m\angle P. In words: the arc \overset{\frown}{CD} looks smaller from the outside point P than from the point A on the circle, by exactly half the near arc.
The chords are perpendicular, so m\angle AEB = 90^\circ. The rule for two chords gives \tfrac12\left(m\overset{\frown}{AB} + m\overset{\frown}{CD}\right) = 90^\circ, so the two arcs add to 180^\circ. All four arcs fill the circle, so \overset{\frown}{BC} and \overset{\frown}{DA} add to 360^\circ - 180^\circ = 180^\circ.
For two tangents, m\angle P = 180^\circ minus the minor arc, so the minor arc is 180^\circ - 60^\circ = 120^\circ. The angles \angle PST and \angle PTS are tangent-chord angles that open toward P, so each intercepts the minor arc and measures \tfrac12 \cdot 120^\circ = 60^\circ. All three angles of triangle PST are 60^\circ, so the triangle is equilateral.