# Thread: Simple (should be!) Geometry Proof: Two tangents to a circle

1. ## Simple (should be!) Geometry Proof: Two tangents to a circle

I found this little proof in a book I was practicing out of. Does anyone have a proof for this? It must be so simple! Gaahh!

Let y be a circle, l and m be two nonparallel lines that are tangent to y at the points P and Q, and let A be the point of intersection of l and m. Prove that:
i) O lies on the bisector of angle(PAQ)
ii) PA = QA
iii) The line PQ is perpendicular to the line OA

I know Triangle POQ is isosceles, and OP = PQ, etc etc all the things about that triangle, but I don't know how to work the point A into the mix, nor do I know anything (I don't think) about Triangle PQA. Any help here?

2. If H is the center of circle: $
\vartriangle OPA \cong OQA
$
, (because AO=OQ (are radios), OA is side in common, and $
\measuredangle HPA = \measuredangle OQA = 90
$
(the radio is perpendicular to the tangent), then for SAS, those triangules are congruent)

for this OA is bisector of <PAQ, and PA=AQ

Now you note that: $
\vartriangle OPQ
$
is isoceles, of base PQ and how $
\measuredangle POA = \measuredangle QOA
$
(congruence) $
\Rightarrow PQ \bot OA
$
for propierty of isoceles triangle

3. Hello, paupsers!

Let $O$ be the center of a circle,
$l$ and $m$ be two nonparallel lines that are tangent to $O$ at the points $P$ and $Q.$
Let $A$ be the point of intersection of $l$ and $m.$

Prove that:

(1) $O$ lies on the bisector of $\angle PAQ$
(2) $PA = QA$
(3) $PQ \perp OA$
Code:
                *    P
*         o
*           /   * o
*           /     *     o
/                 o
*          /        *              o
*       O * - - - - * - - - - - - - - - o A
*          \        *              o
\                 o
*           \     *     o
*           \   * o
*         o
*    Q

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

$OP \perp PA,\;OQ \perp QA$
. .
A radius drawn to the point of tangency is perpencidular to the tangent.

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

$OP = OQ$
. .
In a circle, all radii are equal.

$OA = OA$
. .
Reflexive property

Hence: . $\Delta OPA \cong \Delta OQA$
. .
Side-side theorem for right triangles

Therefore: . $\angle OAP = \angle OAQ$
. .
Corresponding parts of congruent triangles are equal.

(2) Prove: . $PA = QA$
. .
Corresponding parts of congruent triangles are equal.

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

$OP = OQ$ . . .
$PA = QA$ . . .
Hence, $OA$ is the perpendicular bisector of $PQ.$
Therefore: . $PQ \perp OA$