Hello, jenjen!

Here's #3 . . .

3. Let C be a circle with center Q.

Let X be a point in the exterior of C,

and let A and B be two points of C which lie on opposite sides of QX

such that XA and XB are tangent to C.

Prove that: .$\displaystyle XA = XB$ Code:

* * *
* * A
* *
* / * *
/ *
* / * *
* Q* - - - - * - - - *X
* \ * *
\ *
* \ * *
* *
* * B
* * *

Tangent XA is perpendicular to radius QA.

Tangent XB is perpendicular to radius QB.

. . Hence, triangles QAX and QBX are right triangles.

Since QA and QB are radii of circle Q, $\displaystyle QA = QB.$

. . And, of course: .$\displaystyle QX = QX.$

Hence: .$\displaystyle \Delta QAX \cong \Delta QBX$

Therefore: .$\displaystyle XA = XB$