Prove that $HI \perp BD$.

Question

$I$ is the incenter of $\triangle ABC$. The perpendicular bisector of $CI$ cuts $AI$, $BI$ and $CA$ respectively at $D$, $E$ and $F$. The line that passes through the midpoint of $IF$ and is perpendicular to $BC$ intersects the line that passes through $E$ and is perpendicular to $AI$ at $H$. Prove that $IH \perp BD$.

This is supposed to be an easy problem yet I still can't think of a way to solve this. But I have a theory that $H$ is the centre of cyclic quadrilateral $AIFE$, if I could have proved it.

If you can solve the problem, thanks so much for that!

Answer 1

It is not so hard.

First, since $HD=HC$, and $CE$ is the angle bisector, it is trivial that $DHCK$ is a rhombus.

Second, notice that $\angle DGK=\angle GKC-\angle GBC={\angle A\over 2}$.

This means that $\angle BGC=\angle A$, says $G$ is on the circumcirle of $ABC$.

Now since $D$ is the incentre,$AG=GD=GC$.(This indicates the fact that $GJ$ is also the perpendicular bisector of $AD$)

And rhombus contains the fact that $DH$ is parallel to $BC$, means $\angle AHD=\angle C$.

Also, from cyclic, $\angle AGB=\angle C$.

So $AGHD$ is also cyclic. And $J$ is the centre.

Finally, let $JD$ intersect $BF$ at $N$, we have

$$\angle DBN+\angle BDN=\angle DBC+\angle CBF+\angle JDG={\angle B \over 2}+{\angle A \over 2}+{\angle C\over 2} $$

And done.

Answer 2

Your theory is true. First, I assume you proved that $AIFE$ (this is proved by proving that $\angle IAF=\angle IEF=\alpha$).

Now, note that $IF$ is parallel to $BC$(you can easily prove this by proving that $\angle FIC=\angle ICB$). So, $\angle EIF=\angle IBC=\beta$.

Since $AIFE$ is cyclic quadrilateral,we have that $\angle EIF=\angle FAE$. Then you can easily check that $\angle IAE=\angle AIE=\alpha+\beta$.

This proves that $\Delta AEI$ is isosceles and therefore $H$ is in the bisector of$AI$.

Since $H$ is also in the bisector of $IF$, this proves that $H$ is the center of the circle passing through $AIFE$.