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 4 of 5: Inscribed angles force PS parallel to LK

Previous step Next step

\angle SAP=\angle BAK=\angle BLK=\angle PLK=\angle SPB\ \implies\ PS\parallel LK

Detailed analysis

Since $A$ , $P$ , $K$ are colinear (Step 1), the ray $AP$ is the ray $AK$ , so $\angle SAP=\angle SAK$ ; and since $S$ lies on line $AB$ , the ray $AS$ is the ray $AB$ , so $\angle SAK=\angle BAK$ . The inscribed angles $\angle BAK$ and $\angle BLK$ both subtend the same arc $BK$ of $\Gamma$ from the same side, hence $\angle BAK=\angle BLK$ . Since $B$ , $P$ , $L$ are colinear (Step 1), the ray $LB$ is the ray $LP$ , so $\angle BLK=\angle PLK$ . Combining with Step 3, $\angle PLK=\angle SAP=\angle SPB$ . These are the angles that $LK$ and $PS$ make with the common transversal line $BPL$ at $L$ and at $P$ respectively, so equal corresponding angles give $PS\parallel LK$ .