Question

The internal bisector of angle $A$ in a triangle $ABC$ with $AC>AB$, meets the circumcircle $\Gamma$ of the triangle in $D$. Join $D$ to the centre $O$ of the cirlce $\Gamma$ suppose $DO$ meets $AC$ in $E$, possibly when extended. Given that $BE$ is perpendicular to $AD$, show that $AO$ is parallel to $BD$.

Answer 1

Draw an appropriate diagram using the given conditions.

Let $DE$ meet the $BC$ at $X$. Let $AD$ meet the line $BE$ at $Y$ .

As $XE$ is the perpendicular bisector of $BC$, it implies that $BE=EC$. Besides, $AY$ is the angular bisector of $\angle A$ and is perpendicular to the $BE$ implies that $BY=YE$.

If $AY$ bisects $BE$, then $YD$ also bisects $BE$ and is perpendicular to $BE$.

Hence, $YD$ is the angular bisector of $\angle D$.

Now

$\angle BDA=\angle BCA=\angle C$.

Then

$\to$ $\angle ODA=\angle BDA=\angle C$.

As

$OA=OD$

$\to$$\angle ODA=\angle OAD=\angle C$.

Hence

$\to$ $\angle BDA=\angle OAD$.

Thus $BD$ is parallel to $AO$