Problem 2

Let

ABC

be a triangle with circumcenter

O

. The points

P

and

Q

are interior points of the sides

CA

and

AB

, respectively. Let

K

L

, and

M

be the midpoints of the segments

BP

CQ

, and

PQ

, respectively, and let

\Gamma

be the circle passing through

K

L

, and

M

. Suppose that the line

PQ

is tangent to the circle

\Gamma

. Prove that

OP = OQ

Step 4 of 4: Finish with power of a point

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In plain words

Q and P lie on chords AB and CA of the circumcircle, so their power with respect to it is exactly the product just computed.

OA^2-OQ^2=AQ\cdot QB=AP\cdot PC=OA^2-OP^2 \implies OP=OQ

Detailed analysis

Since $Q$ lies on chord $AB$ of the circumcircle of $ABC$ (center $O$ , radius $OA$ ), the power of $Q$ gives $AQ\cdot QB=OA^2-OQ^2$ . Since $P$ lies on chord $CA$ , similarly $AP\cdot PC=OA^2-OP^2$ . Step 3 showed $AQ\cdot QB=AP\cdot PC$ , so $OA^2-OQ^2=OA^2-OP^2$ , giving $OP^2=OQ^2$ and hence $OP=OQ$ . $\blacksquare$