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 1 of 5: Read off the collinearities and introduce X

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In plain words

K, L, M are just the second intersections of the cevians AP, BP, CP with the circle, so each cevian is literally a chord through P; a second tangent from M gives a fresh isosceles triangle to compare with SPC.

A,P,K\ \text{colinear};\quad B,P,L\ \text{colinear};\quad C,P,M\ \text{colinear};\qquad X=(\text{tangent at }M)\cap SC

Detailed analysis

By definition $K$ is the second intersection of line $AP$ with $\Gamma$ , so $A$ , $P$ , $K$ are colinear; likewise $B$ , $P$ , $L$ are colinear and $C$ , $P$ , $M$ are colinear, with $P$ strictly between $C$ and $M$ (since $P$ is interior to the triangle, hence interior to $\Gamma$ ). Let the tangent to $\Gamma$ at $M$ meet the line $SC$ (which carries the tangent to $\Gamma$ at $C$ , since $S$ was defined on that tangent) at a point $X$ . This $X$ is the vertex from which we will compare two isosceles triangles.