Question 49
Let $G$, $H$ be the centroid and orthocentre of $\triangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot ABC$ again at $L \neq A$. The tangent to $\omega$ at $L$ intersects $\odot ABC$ at $K \neq L$.
Given that $AG = GH$, prove $\angle HKG = 90^\circ$.
Given that $AG = GH$, prove $\angle HKG = 90^\circ$.