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 1 of 4: Identify K, L, M as midline endpoints

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In plain words

M is the midpoint of both triangles PQB and QPC together with K, L, so MK and ML are midlines parallel to the outer sides.

MK \parallel AB,\ MK=\tfrac12 QB \qquad ML \parallel AC,\ ML=\tfrac12 PC

Detailed analysis

In triangle $PQB$ , $M$ and $K$ are the midpoints of $PQ$ and $PB$ , so $MK$ is a midline: $MK\parallel QB$ , i.e. $MK\parallel AB$ , with $MK=\tfrac12 QB$ . Likewise in triangle $QPC$ , $M$ and $L$ are the midpoints of $QP$ and $QC$ , so $ML\parallel PC$ , i.e. $ML\parallel AC$ , with $ML=\tfrac12 PC$ . Since $MK\parallel AB$ and $ML\parallel AC$ , the angle between $MK$ and $ML$ equals the angle between $AB$ and $AC$ : $\angle KML=\angle BAC$ . Also $\dfrac{ML}{MK}=\dfrac{PC}{QB}$ .