Angles Inside and Outside Circles

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Learning goals

Identify the arc endpoints for inside, on-circle, and outside angles
Justify the intersecting-chord half-sum using inscribed angles and a triangle
Derive exterior angle rules for secants and tangents from earlier angle facts
Calculate angles and missing arcs while keeping the chosen rays fixed
Reconstruct a full circle from linked inside and outside information

Locate the vertex before choosing arcs

An angle inside a circle combines two arcs. An angle outside compares two arcs. An angle on the circle uses one arc, as in Inscribed Angles.

These are separate, matched configurations. They are schematic, not successive frames of one moving construction. Every named circle point is distinct, and every angle here is an ordinary angle between specified rays, between $0^\circ$ and $180^\circ$ .

Inside: two chords cross

Inside. AC and BD cross at P. Select arcs AB and CD, facing the angle APB and its vertical angle.

If arcs $AB$ and $CD$ measure $70^\circ$ and $110^\circ$ , their half-angles add to $90^\circ$ . The other two consecutive arcs together measure $180^\circ$ , completing the circle.

On: one intercepted arc

On the circle. The arc AB selected by angle AVB excludes V.

An inscribed angle intercepting an arc of $140^\circ$ measures $70^\circ$ . Its vertex must be excluded from that arc; the other arc completes $360^\circ$ .

Outside: two rays point into the circle

Outside. Along one ray the order is P,A,B; along the other it is P,D,C. Select far arc BC and near arc DA.

The near arc joins the first intersections $A,D$ without passing through $B,C$ . The far arc joins the second intersections $B,C$ without passing through $A,D$ . These names come from ray order, not from left and right on the page.

For far arc $BC$ of $158^\circ$ and near arc $DA$ of $42^\circ$ , the difference of their half-angles is $58^\circ$ . The other two consecutive arcs total $160^\circ$ , so the selected arcs fit one full circle.

Why intersecting chords use a half-sum

Let $A,B,C,D$ occur in that order around a circle. Chords $AC$ and $BD$ cross at an interior point $P$ . For $\angle APB$ , use arc $AB$ excluding $C,D$ and arc $CD$ excluding $A,B$ . They face this angle and its vertical angle, not its adjacent angle.

Join A to D. Angle APB is an exterior angle of triangle APD because PB is opposite PD.

The intersecting-chord angle rule#

With arc $AB$ of $82^\circ$ and arc $CD$ of $116^\circ$ , the inscribed angles $PDA$ and $PAD$ measure $41^\circ$ and $58^\circ$ . The exterior-angle theorem adds them, giving $99^\circ$ . The remaining two arcs total $162^\circ$ and may each be $81^\circ$ .

Write $a$ for the degree measure of the selected arc $AB$ , and $c$ for that of $CD$ . Since $P$ lies on $AC$ and $BD$ , the angle $PAD$ is the inscribed angle $CAD$ , and $PDA$ is the inscribed angle $BDA$ .

\begin{aligned} m\angle PAD&=c/2,\\ m\angle PDA&=a/2. \end{aligned}

Ray $PB$ extends side $PD$ of triangle $APD$ . The earlier triangle exterior-angle theorem gives

\begin{aligned} m\angle APB&=a/2+c/2,\\ m\angle APB&=(a+c)/2. \end{aligned}

Nothing depended on $82$ or $116$ . The proof requires two chords crossing strictly inside, with the selected arcs facing a vertical pair. Applying the same argument to $\angle BPC$ uses the other two arcs and covers the adjacent, possibly acute or obtuse, angle.

Worked example 1 Choose the vertical pair, then its neighbor

Problem. Points $A,B,C,D$ occur in order around a circle. Chords $AC$ and $BD$ meet at $P$ . The consecutive arcs have the following measures. Find $m\angle APB$ and $m\angle BPC$ .

\begin{gathered} m(\widehat{AB})=78^\circ,\\ m(\widehat{BC})=64^\circ,\\ m(\widehat{CD})=132^\circ,\\ m(\widehat{DA})=86^\circ. \end{gathered}

The highlighted arcs belong to APB and its vertical angle. All four givens are stated above.

Solution. The positive arcs total $360^\circ$ . Angle $APB$ uses $AB$ and $CD$ .

\begin{aligned} m\angle APB&=(78^\circ+132^\circ)/2,\\ m\angle APB&=105^\circ. \end{aligned}

The adjacent angle uses $BC$ and $DA$ .

\begin{aligned} m\angle BPC&=(64^\circ+86^\circ)/2,\\ m\angle BPC&=75^\circ. \end{aligned}

Their sum is $180^\circ$ , matching the straight angle along $AC$ .

Check your understanding

Chords $AC$ and $BD$ cross at interior $P$ , with circular order $A,B,C,D$ . Arc $AB$ measures $55^\circ$ and arc $CD$ measures $101^\circ$ . Find $m\angle APB$ .

78^\circ

28^\circ

156^\circ

102^\circ

Why two secants use a half-difference

Let $P$ be strictly outside, with ray orders $P,A,B$ and $P,D,C$ . The four circle points occur in order $A,B,C,D$ . Join $B$ to $D$ .

Triangle PBD uses the near point D and far point B. Ray DC extends DP.

The two-secant exterior angle rule#

Take far arc $BC$ of $172^\circ$ and near arc $DA$ of $36^\circ$ . Inscribed angle $BDC$ is $86^\circ$ ; inscribed angle $PBD$ , also $ABD$ , is $18^\circ$ . The exterior angle $BDC$ equals the angle at $P$ plus the angle at $B$ , so the angle at $P$ is $68^\circ$ . The other arcs may each measure $76^\circ$ , making a positive full circle.

Let the far arc $BC$ have degree measure $f$ , and the near arc $DA$ have measure $n$ . The inscribed-angle theorem gives

\begin{aligned} m\angle BDC&=f/2,\\ m\angle PBD&=n/2. \end{aligned}

Because $P,D,C$ are in that order, $BDC$ is exterior to triangle $PBD$ .

\begin{aligned} f/2&=m\angle APD+n/2,\\ m\angle APD&=(f-n)/2. \end{aligned}

The numbers $172$ and $36$ only illustrated these steps. The ray orders force the far arc to exceed the near arc. The rule measures the angle between the two rays pointing toward the circle. Reversing just one ray gives its supplement, not the same angle.

The far arc can be major. Name its route explicitly rather than assuming every two-letter arc is minor. In the formulas, $f$ always names the selected route between the far endpoints, even when that route exceeds $180^\circ$ .

Worked example 2 A major far arc

Problem. Two secants have orders $P,A,B$ and $P,D,C$ . Around the circle, the consecutive arcs $AB,BC,CD,DA$ measure $44^\circ,206^\circ,60^\circ,50^\circ$ . Arc $BC$ here means the route excluding $A,D$ . Identify the near and far arcs and find $m\angle APD$ .

The near endpoints are A,D, and the far endpoints are B,C. The diagram is schematic, including arc sizes.

Solution. Near arc $DA$ excludes $B,C$ and measures $50^\circ$ . Far arc $BC$ excludes $A,D$ and is a major arc of $206^\circ$ . All four positive arcs total $360^\circ$ .

\begin{aligned} m\angle APD&=(206^\circ-50^\circ)/2,\\ m\angle APD&=78^\circ. \end{aligned}

Check your understanding

Secant ray orders are $P,A,B$ and $P,D,C$ . The far arc $BC$ excluding $A,D$ measures $164^\circ$ ; near arc $DA$ excluding $B,C$ measures $38^\circ$ . Find $m\angle APD$ .

101^\circ

126^\circ

63^\circ

82^\circ

A tangent and a secant

The same subtraction works when one ray is tangent. Let $PT$ touch at $T$ and let the secant order be $P,A,B$ . Near arc $TA$ excludes $B$ . Far arc $BT$ excludes $A$ . The third arc is the route from $A$ to $B$ excluding $T$ .

Join T to A. The tangent ray TP selects near arc TA; angle TAB selects far arc BT.

The tangent-secant exterior angle rule#

For near arc $TA$ of $48^\circ$ and far arc $BT$ of $168^\circ$ , the tangent-chord theorem gives $24^\circ$ at $T$ . Inscribed angle $TAB$ is $84^\circ$ . Since $AP$ is opposite $AB$ , this is exterior to triangle $PTA$ , leaving $60^\circ$ at $P$ . The remaining arc $AB$ is $144^\circ$ .

Let near arc $TA$ have degree measure $n$ , and far arc $BT$ have measure $f$ , using the routes just specified. The tangent-chord theorem for ray $TP$ and chord $TA$ gives $n/2$ at $T$ . The inscribed-angle theorem gives $f/2$ for angle $TAB$ .

Angle $TAB$ is exterior to triangle $PTA$ , so

\begin{aligned} f/2&=m\angle TPA+n/2,\\ m\angle TPA&=(f-n)/2. \end{aligned}

The numerical argument used these same ray choices. Reflecting the picture changes no step. A major far arc makes $TAB$ obtuse, but the exterior-angle reasoning still applies. Strict exterior $P$ , a genuine tangent at $T$ , and two distinct secant intersections are required.

Worked example 3 An obtuse tangent-secant angle

Problem. Segment $PT$ is tangent at $T$ , and a secant has order $P,A,B$ . Near arc $TA$ excluding $B$ measures $40^\circ$ . Far arc $BT$ excluding $A$ measures $226^\circ$ . Find $m\angle TPA$ and $m\angle PAT$ , and check the remaining arc.

The secant meets A before B. The dotted chord TA completes triangle PTA.

Solution. The remaining arc $AB$ excluding $T$ is $94^\circ$ , positive. The three selected arcs total $360^\circ$ .

\begin{aligned} m\angle TPA&=(226^\circ-40^\circ)/2,\\ m\angle TPA&=93^\circ. \end{aligned}

Angle $TAB$ intercepts the far arc and is $113^\circ$ . Its supplement is the requested angle at $A$ .

\begin{aligned} m\angle PAT&=180^\circ-113^\circ,\\ m\angle PAT&=67^\circ. \end{aligned}

The angle at $T$ is $20^\circ$ ; the triangle check is $93^\circ+67^\circ+20^\circ=180^\circ$ .

Check your understanding

With tangent $PT$ and secant order $P,A,B$ , near arc $TA$ excluding $B$ is $32^\circ$ and far arc $BT$ excluding $A$ is $244^\circ$ . Find $m\angle TPA$ .

138^\circ

106^\circ

122^\circ

74^\circ

Two tangents and both angle sizes

Let rays $PT,PU$ touch the same circle at distinct $T,U$ , with $P$ strictly outside. The near arc is the minor arc $TU$ facing $P$ ; choose $R$ on the other arc to name the far arc $TRU$ .

An acute angle between tangent rays. The dotted contact chord TU makes triangle PTU.

If the minor arc is $118^\circ$ , each tangent-chord base angle of triangle $PTU$ is $59^\circ$ . The angle at $P$ is $62^\circ$ . The far arc is $242^\circ$ , whose excess over the near arc is $124^\circ$ , twice that angle.

The two-tangent exterior angle rule#

First use a minor arc of $134^\circ$ . Both tangent-chord base angles of triangle $PTU$ measure $67^\circ$ , leaving $46^\circ$ at $P$ . The major arc is $226^\circ$ , and half its difference from $134^\circ$ is also $46^\circ$ .

Now let the minor arc measure $n$ , strictly between $0^\circ$ and $180^\circ$ . The two contact points are distinct and not diametrically opposite: tangents at opposite diameter endpoints would be parallel and could not meet at finite $P$ . Both tangent-chord base angles select the near arc and measure $n/2$ .

The triangle sum gives

\begin{aligned} m\angle TPU&=180^\circ-n,\\ m(\widehat{TRU})&=360^\circ-n,\\ \frac{(360^\circ-n)-n}{2}&=180^\circ-n. \end{aligned}

Thus the two-tangent angle is half the far-minus-near arc difference too. The choice of $134^\circ$ was unnecessary. The angle is acute when $n>90^\circ$ , right when $n=90^\circ$ , and obtuse when $n<90^\circ$ .

Worked example 4 Recover both arcs from an obtuse angle

Problem. Rays $PT$ and $PU$ are tangent to one circle at distinct points. Given $m\angle TPU=104^\circ$ , find the minor arc $TU$ and major arc $TRU$ , where $R$ is on the major arc.

The tangent rays form an obtuse angle. Neither arc is labeled with a computed measure.

Solution. The near minor arc and the angle at $P$ sum to $180^\circ$ .

\begin{aligned} m(\widehat{TU})&=180^\circ-104^\circ,\\ m(\widehat{TU})&=76^\circ. \end{aligned}

The other arc completes the circle.

\begin{aligned} m(\widehat{TRU})&=360^\circ-76^\circ,\\ m(\widehat{TRU})&=284^\circ. \end{aligned}

Both arcs are positive, the minor arc is below $180^\circ$ , and half their difference is the given $104^\circ$ .

Check your understanding

Tangents from strict external $P$ touch a circle at $T,U$ . Minor arc $TU$ is $68^\circ$ . Find the angle between rays $PT$ and $PU$ .

34^\circ

68^\circ

146^\circ

112^\circ

Reconstruct the whole circle

Use the vertex position to choose a rule, name its arc routes, and only then write an equation. When several rules share one circle, finish by checking every consecutive arc is positive and their sum is $360^\circ$ .

Worked example 5 Link an inside sum to an outside difference

Problem. Points $A,B,C,D$ occur in order around a circle. Chords $AC,BD$ meet at $X$ . Lines $AB,DC$ meet at external $P$ with orders $P,A,B$ and $P,D,C$ . The angle $AXB$ measures $92^\circ$ , the angle $APD$ measures $36^\circ$ , and consecutive arc $AB$ measures $81^\circ$ . Find the other three consecutive arc measures. Here arc $BC$ excludes $A,D$ and arc $DA$ excludes $B,C$ .

X is the chord intersection. P is a different, external intersection of the side lines.

Solution. The inside angle uses arcs $AB,CD$ . Double it before subtracting arc $AB$ .

\begin{aligned} m(\widehat{CD})&=2(92^\circ)-81^\circ,\\ m(\widehat{CD})&=103^\circ. \end{aligned}

The remaining arcs $BC,DA$ total $176^\circ$ . Their difference is twice the exterior angle, $72^\circ$ . Let their degree measures be $b,d$ .

\begin{aligned} b+d&=176,\\ b-d&=72,\\ 2b&=248,\\ b&=124,\\ d&=52. \end{aligned}

The four positive arc measures are $81^\circ,124^\circ,103^\circ,52^\circ$ , totaling $360^\circ$ . The far arc exceeds the near arc, as the specified secant orders require. Both original angle equations hold.

Check your understanding

In the linked configuration of Example 5, use new givens: angle $AXB$ measures $83^\circ$ and angle $APD$ measures $29^\circ$ . What is the near arc $DA$ excluding $B,C$ ?

126^\circ

97^\circ

68^\circ

14^\circ

Angles Inside and Outside Circles

Learning goals

Locate the vertex before choosing arcs

Inside: two chords cross

On: one intercepted arc

Outside: two rays point into the circle

Why intersecting chords use a half-sum

Why two secants use a half-difference

A tangent and a secant

Two tangents and both angle sizes

Reconstruct the whole circle

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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