Problem 4

Let

P

be a point interior to triangle

ABC

(with

CA \neq CB

). The lines

AP

BP

and

CP

meet again its circumcircle

\Gamma

K

L

, respectively

M

. The tangent line at

C

\Gamma

meets the line

AB

S

. Show that from

SC = SP

follows

MK = ML

Step 2 of 5: Two isosceles triangles force MX parallel to PS

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XM=XC,\ SP=SC\ \implies\ \angle XMC=\angle XCM=\angle SCP=\angle SPC\ \implies\ MX\parallel PS

Detailed analysis

Since $XM$ and $XC$ are the two tangent segments from $X$ to $\Gamma$ , they are equal, so triangle $XMC$ is isosceles and $\angle XMC=\angle XCM$ . Because $C$ , $P$ , $M$ are colinear (Step 1), the ray $CM$ is the ray $CP$ , so $\angle XCM=\angle XCP$ ; and because $X$ lies on line $SC$ , the ray $CX$ is the ray $CS$ , so $\angle XCP=\angle SCP$ . On the other hand, $SP=SC$ is given, so triangle $SPC$ is isosceles with apex $S$ , giving $\angle SPC=\angle SCP$ . Chaining these equalities, $\angle XMC=\angle SPC$ . These are the angles that $MX$ and $PS$ make with the common transversal line $CPM$ at $M$ and at $P$ respectively, so equal corresponding angles give $MX\parallel PS$ .