Question

Let $ABC$ be triangle in which $AB=AC$. Suppose the orthocentre of the triangle lies on the incircle. Find the ratio $\frac{AB}{BC}$.

• A. $\frac{4}{3}$
• B. $\frac{3}{5}$
• C. $\frac{3}{4}$
• D. $\frac{5}{3}$

Answer 1

The correct option is C $\frac{3}{4}$

Let $D$ be the mid-point of $BC$.

Extend $AD$ to meet the circumcircle in $L$.

Then we know that $HD=DL$.

But $HD=2r$.

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, $△BLD$ is similar to $△ABD$.

So, $\frac{AB}{BD}=\frac{BL}{LD}=\frac{3}{2}$

Finally,

$\frac{AB}{BC}=\frac{AB}{2BD}=\frac{3}{4}$