answers: question 6

the question was

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$ ). The lines $AP$ , $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$ , $L$ , respectively $M$ . The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$ . Show that from $SC = SP$ follows $MK = ML$ .

solution

Without loss of generality, suppose that $AS > BS$ . By Power of a Point, $SP^2 = SC^2 = SB \cdot SA$ , so $\overline{SP}$ is tangent to the circumcircle of $\triangle ABP$ . Thus, $\angle KPS = 180 - \angle SPA = \widehat{AP}/2 = \angle ABP$ . It follows that after some angle-chasing,
$\begin{align*}\widehat{ML} &= \widehat{MA} + 2\angle AKL \\ &= \left(2\angle CPK - \widehat{KC}\right) + 2\angle ABL ...$ so $ML = MK$ as desired.

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Solution 2

Let the tangent at $M$ to $\Gamma$ intersect $SC$ at $X$ . We now have that since $\triangle{XMC}$ and $\triangle{SPC}$ are both isosceles, $\angle{SPC}=\angle{SCP}=\angle{XMC}$ . This yields that $MX \| PS$ .

Now consider the power of point $S$ with respect to $\Gamma$ .

$SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}$

Hence by AA similarity, we have that $\triangle{SPA} \sim \triangle{SBP}$ . Combining this with the arc angle theorem yields that $\angle{SPA}=\angle{SBP}=\angle{PKL}$ . Hence $PS \| LK$ .