# Circle Theorems

• Nov 10th 2008, 09:41 AM
bearej50
Circle Theorems
Given: Isosceles Triangle ABC is inscribed in a circle with base BC.
Prove: If P is any poin on minor arc BC, then ray PA bisects angle BPC.
• Nov 10th 2008, 11:58 AM
Soroban
Hello, bearej50!

Quote:

Given: Isosceles Triangle $\displaystyle ABC$ is inscribed in a circle with base $\displaystyle BC.$

Prove: If $\displaystyle P$ is any point on minor arc $\displaystyle BC$, then ray $\displaystyle PA$ bisects $\displaystyle \angle BPC.$

Code:

                A               * o *           *    / \    *         *    /  \    *       *    /  o \    *             /      \       *    /        \    *       *  /      o  \  *       *  /            \  *         /              \       B*- - - - - -o- - -*C         *              *           *          *               * * * o                     P

Draw chords $\displaystyle PB$ and $\displaystyle PC$.

Inscribed angle $\displaystyle APB$ is measured by $\displaystyle \tfrac{1}{2}\,\text{arc}(AB).$
Inscribed angle $\displaystyle APC$ is measured by $\displaystyle \tfrac{1}{2}\,\text{arc}(AC).$

Since $\displaystyle AB = AC$ .
(the triangle is isosceles),
. . then $\displaystyle \text{arc}(AB) = \text{arc}(AC)$ .
(equal chords subtend equal arcs).

Therefore: .$\displaystyle \angle APB = \angle APC \quad\Rightarrow\quad AP \text{ bisects }\angle BPC.$

• Nov 10th 2008, 04:56 PM
bearej50
thank you very much!