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

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• Apr 21st 2009, 03:54 PM
paupsers
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?
• Apr 21st 2009, 04:12 PM
Nacho
If H is the center of circle: \$\displaystyle
\vartriangle OPA \cong OQA
\$, (because AO=OQ (are radios), OA is side in common, and \$\displaystyle
\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: \$\displaystyle
\vartriangle OPQ
\$ is isoceles, of base PQ and how \$\displaystyle
\measuredangle POA = \measuredangle QOA
\$ (congruence) \$\displaystyle
\Rightarrow PQ \bot OA
\$ for propierty of isoceles triangle
• Apr 21st 2009, 04:52 PM
Soroban
Hello, paupsers!

Quote:

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\$