Hello, paupsers!

Let $\displaystyle O$ be the center of a circle,

$\displaystyle l$ and $\displaystyle m$ be two nonparallel lines that are tangent to $\displaystyle O$ at the points $\displaystyle P$ and $\displaystyle Q.$

Let $\displaystyle A$ be the point of intersection of $\displaystyle l$ and $\displaystyle m.$

Prove that:

(1) $\displaystyle O$ lies on the bisector of $\displaystyle \angle PAQ$

(2) $\displaystyle PA = QA$

(3) $\displaystyle PQ \perp OA$ Code:

* P
* o
* / * o
* / * o
/ o
* / * o
* O * - - - - * - - - - - - - - - o A
* \ * o
\ o
* \ * o
* \ * o
* o
* Q

(1) Prove: $\displaystyle OA$ bisects $\displaystyle \angle PAQ.$

$\displaystyle OP \perp PA,\;OQ \perp QA$

. . A radius drawn to the point of tangency is perpencidular to the tangent.

Hence, $\displaystyle \Delta OPA$ and $\displaystyle \Delta OQA$ are right triangles.

$\displaystyle OP = OQ$

. . In a circle, all radii are equal.

$\displaystyle OA = OA$

. . Reflexive property

Hence: .$\displaystyle \Delta OPA \cong \Delta OQA$

. . Side-side theorem for right triangles

Therefore: .$\displaystyle \angle OAP = \angle OAQ$

. . Corresponding parts of congruent triangles are equal.

(2) Prove: .$\displaystyle PA = QA$

. . Corresponding parts of congruent triangles are equal.

(3) Prove: .$\displaystyle PQ \perp OA$

$\displaystyle OP = OQ$ . . . They are radii.

$\displaystyle PA = QA$ . . . proved in (1)

Hence, $\displaystyle OA$ is the perpendicular bisector of $\displaystyle PQ.$

. . Two points equidistant from the ends of a line segment determine the perpendicular bisector of the line segment.

Therefore: .$\displaystyle PQ \perp OA$