Question

The altitudes from the vertices A, B and C of the triangle ABC meet it circumcircle at D, E

Answer 1

$\angle FDA=\angle FCA=\frac{\pi }{2}-A$

$\angle ADE=\angle ABE=\frac{\pi }{2}-A$

$\Rightarrow \angle FDA=\pi -2A$

Similarly, $\angle DFE=\pi -2C$

$\angle DEF=\pi -2B$

The angles of triangle DEF are $\pi -2A$, $\pi -2B$ and $\pi -2C$ respectively.

Also, it is given that $\frac{z_3-z_1}{z_2-z_1}$ is purely real.

Hence, $arg\left(\frac{z_3-z_1}{z_2-z_1}\right)=0or\pi$

$\Rightarrow \pi -2A=0or\pi$

$\Rightarrow A=\frac{\pi }{2}or0$ (not permissible)

Hence triangle ABC is right angle at A.

Hence proved.