Let $ABC$ be a triangle for which $AB = AC$. Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio $\frac{AB}{BC}$.
This problem is taken from the $31$st Indian National Mathematical Olympiad.
I am looking for all possible solutions to this problem( to better understand $- $ the strategies to solve Geometry problems)
Note:- I don't know much Co-ordinate Geometry. So, please restrict your answers to the scope of Euclidean Geometry.
Since the triangle is isosceles, the orthocentre lies on the perpendicular $AD$ from $A$ on to $BC$. Let it cut the incircle at $H$.
Let $D$ be the mid-point of $BC$. Extend $AD$ to meet the circumcircle in $L$. Then we know that $HD=DL$. But, we have, $HD=2r$ where $r$ is the inradius of triangle $ABC$. Thus, $DL=2r$. Therefore, $IL=ID+DL=r+2r=3r$. We also know that $LB = LI$. Therefore $LB = 3r$. This gives $$\frac{BL}{LD} = \frac{3r}{2r} = \frac{3}{2}$$ But triangle $BLD$ is similar to triangle $ABD$. So Finally, we get, $$\frac{AB}{BD} = \frac{BL}{LD} = \frac{3}{2}$$ Hence, $$\frac{AB}{BC} = \frac{AB}{2BD} = \frac{3}{4}$$ Hope it helps.