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 2 of 4: Chase the tangent-chord angle at M

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\angle AQP=\angle KMQ=\angle MLK

Detailed analysis

Since $MK\parallel AQ$ (part of line $AB$ ), the transversal $QM$ gives equal alternate angles $\angle AQP=\angle AQM=\angle KMQ$ (as $M$ lies on segment $PQ$ , ray $QM$ is ray $QP$ ). Since $PQ$ is tangent to $\Gamma$ at $M$ , the tangent–chord angle theorem applied to chord $MK$ gives $\angle KMQ=\angle MLK$ , the inscribed angle subtending $MK$ from $L$ in the alternate segment. Hence $\angle AQP=\angle MLK$ .