In triangle ABC, diameter and radius of circumcircle meet BC at Q and M. Prove $\frac{[AQC]}{[MTC]}=\left(\frac{\sin B}{\cos C}\right)^2$

Question

In triangle ABC, the height from A cuts the circumcircle at T. The diameter of the circle that passes through A and the radius OT cut the side BC at Q and M, respectively. Demonstrate that $$\frac{[AQC]}{[MTC]}=\bigg(\frac{\sin B}{\cos C}\bigg)^2$$ I draw this figure And working with the triangles I found that $$[AQC]=R^2\cos2B\sin2B\\ [MTC]=4R^2\sin2B\sin^3B$$ But, how can I find the relation of this areas with the angle C? Can someone help me?

Thanks for attention.

Answer 1

Note

\begin{align} & \angle BCT = \angle BAT = 90-\angle B \\ & \angle CAQ = \angle ACO = \frac12(180 - \angle AOC) = 90-\angle B\\ \end{align}

which leads to $ \angle BCT = \angle CAQ$, $\angle CTO = \angle OCT= \angle ACB $ and, hence, similar triangles CMT and CAQ. Thus,

$$\frac{[AQC]}{[MTC]}= \left( \frac { AC}{CT} \right)^2= \left( \frac { \sin \angle CTA}{\sin \angle CAT} \right)^2= \left(\frac{\sin B}{\sin(90- C)}\right)^2 = \left(\frac{\sin B}{\cos C}\right)^2 $$