Show that the center of the circle $(ABC) $ is on $AE $.

Question

Let $\triangle ABC $ and $D\in [BC] $ s.t. $\angle BAD=\angle DAC $.

Let $BE\perp AD $ where $E $ is on the circle (ABD).

Show that the center of the circle $(ABC) $ is on $AE $.

I have no idea how to start.

Answer 1

Angle chasing is enough for this problem:

$$\angle ABE = 90^\circ - \angle BAD = 90^\circ - \frac A2,$$ where $A$ denotes the angle at $A$ of $\triangle ABC$. Now $$ \begin{aligned}\angle DAE &= \angle BDE = B - \angle ABE\\ &= B - (90^\circ - A/2)\\ &= B + A/2 - 90^\circ \end{aligned}$$ so $$ \begin{aligned} \angle BAE &= A/2 + \angle DAE\\ &= B + A - 90^\circ\\ &= 90^\circ - C. \end{aligned}$$

It remains to show that $\angle BAO = 90^\circ - C$ which is obvious.

Note: Of course we need to consider the case where $C > 90^\circ$, but it would be very similar.