Question 75
Let $\Gamma$ be a circle and $P$ a point outside $\Gamma$. The tangents to $\Gamma$ from $P$ touch $\Gamma$ at $A$ and $B$. Let $K$ be a point distinct from $A$ and $B$ on the segment $[AB]$. The circumcircle of triangle $PBK$ meets the circle $\Gamma$ again at the point $T$. Let $P'$ be the reflection of the point $P$ with respect to the point $A$. Prove that $\widehat{\mathrm{PBT}} = \widehat{\mathrm{P}'\mathrm{KA}}$.