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 5 of 5: Conclude M bisects arc LK, so MK = ML

Previous step Next step

MX\parallel PS\parallel LK\ \implies\ MX\parallel LK\ \implies\ M\ \text{is the midpoint of arc } LK\ \implies\ MK=ML

Detailed analysis

Steps 2 and 4 give $MX\parallel PS$ and $PS\parallel LK$ ; by transitivity of parallelism, $MX\parallel LK$ . But $MX$ is (part of) the tangent to $\Gamma$ at $M$ , and $LK$ is a chord of $\Gamma$ : a tangent at a point of a circle is parallel to a chord exactly when that point is the midpoint of one of the two arcs the chord determines. Hence $M$ is the midpoint of arc $LK$ (not containing the other intersection with line $CM$ ), which means the arcs $MK$ and $ML$ are equal, so the chords they subtend are equal: $MK=ML$ . $\blacksquare$