MathGoldMedalist

Home Toolkit Problems Competitions About

Sign In Register

Home Toolkit Problems Competitions About

Home
›
Toolkit
›
Angles in Circles

Angles in Circles

Lesson · Intermediate

Geometry: Plane Geometry

Central Angle

Central angle and its intercepted arc

Inscribed Angle

Inscribed angle and its intercepted arc

Tangent-Chord Angle

Tangent-chord angle and its intercepted arc

Interior Angle

Angle formed by two chords intersecting inside a circle

Exterior Angle

Angle formed by two secants intersecting outside a circle

\angle PAD=\frac{\widehat{CD}}{2}=\frac{\alpha}{2}

\angle PDA=\frac{\widehat{AB}}{2}=\frac{\beta}{2}

Interior angle proof with intersecting chords

$\angle DPC$ is the exterior angle of $\triangle ADP$

\begin{aligned} \angle DPC &=\angle PAD+\angle PDA\\ &=\frac{\alpha}{2}+\frac{\beta}{2}\\ &=\frac{\alpha+\beta}{2} \end{aligned}

Interior angle equals half the sum of the intercepted arcs

\angle BDC=\frac{\widehat{BC}}{2}=\frac{\alpha}{2}

\angle DCE=\frac{\widehat{DE}}{2}=\frac{\beta}{2}

Exterior angle proof with two secants

$\angle DCE$ is the exterior angle of $\triangle ACD$

\angle DCE=\angle CAD+\angle BDC

\frac{\beta}{2} = \angle EAD+\frac{\alpha}{2}

\angle EAD=\frac{\beta-\alpha}{2}

Exterior angle equals half the difference of the intercepted arcs

$B,D,$ and $C$ are collinear. $PB$ and $PC$ are tangents to $\omega_1$ and $\omega_2$ , respectively. Find $\angle BPC+\angle BAC$ .

Two intersecting circles with tangents PB and PC

First angle relation in the two-circle configuration

\angle BAD=\frac{\widehat{BD}}{2}=\angle PBD

Let

\angle BAD=\angle PBC=\alpha

Angle relations alpha and beta in the two-circle configuration

Similarly,

\angle PCB=\angle DAC=\beta

In $\triangle PBC$ ,

\angle BPC = 180^\circ-\alpha-\beta = 180^\circ-\angle BAC

Therefore,

\boxed{\angle BPC+\angle BAC=180^\circ}

AMC 10B 2025 (Problem 5)

Basic Rules of Cyclic Quadrilaterals

← Back to Toolkit

Home Toolkit Problems Competitions About

Stay Connected

Sign up for updates and new content.

Email address

© 2026 MathGoldMedalist. All rights reserved.