# Prove that $HI \perp BD$.

$$I$$ 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!

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.

• I know right. I was about to start typing the answer when yours came in, so I read everything it matches (except a few small notes). – Lê Thành Đạt Apr 30 '19 at 15:43
• @LêThànhĐạt Haha, good for you – StAKmod Apr 30 '19 at 15:44
• I forgot to mark your answer as the correct one. Good job. – Lê Thành Đạt Apr 30 '19 at 15:47

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 $$AIFE$$.