In $\Delta ABC$ , bisectors of $\angle A, \angle B$ meet at incentre $I$. The extension of $AI$ meets the circumcircle of $\Delta ABC$ at $D$.

Question

In $\Delta ABC$ , bisectors of $\angle A, \angle B$ meet at incentre $I$. The extension of $AI$ meets the circumcircle of $\Delta ABC$ at $D$. Let $P$ be the foot of the perpendicular from $B$ to $AD$, and $Q$ be a point on the extension of $AD$ such that $ID = IQ$. Find $\frac{(BQ * IB)}{(BP * ID)}$.

What I Tried: Here is a picture :-

I was struggling to make a good picture out of this, and I am now totally stumped to solve the problem. A good amount of information is not given, for example no lengths of any sides of the triangles or circles are given, and they expect me to find the value of $\frac{(BQ * IB)}{(BP * ID)}$ . Second is that I think the problem looks very complicated so I cannot understand how I should start solving it.

Can anyone help me? Thank You.

Answer 1

It suffices to prove that $DI = DB$.

Note that $\angle IBC = \angle IBA$ and $\angle BAI = \angle CAI = \angle CBD$.

Hence $\angle DBI = \angle DBC + \angle CBI = \angle IAB + \angle IBA = \angle DIB$.

This proves $DI = DB = DQ$. By Thales' Theorem (or otherwise), $IB \perp BQ$.

Now consider the areas of $\triangle DBI$ and $\triangle QBI$ and you will have the ratio you need.