
5. Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Let $E$ and $F$ be respectively the mid-points of $A B$ and $A C$. Suppose the incentre $I$ of triangle $A B C$ lies on the circumcircle of triangle $A E F$. Find the ratio $B C / A B$.


Solution: Draw $I D \perp A C$. Then $I D=r$, the inradius of $\triangle A B C$. Observe $E F \| B C$ and hence $\angle A E F=\angle A B C=90^{\circ}$. Hence $\angle A I F=90^{\circ}$. Therefore $I D^{2}=F D \cdot D A$. If $a>c$, then $F A>D A$ and we have

$$
D A=s-a, \quad \text { and } \quad F D=F A-D A=\frac{b}{2}-(s-a)
$$

Thus we obtain

$$
r^{2}=\frac{(b+c-a)(a-c)}{4}
$$

But $r=(c+a-b) / 2$. Thus we obtain

$$
(c+a-b)^{2}=(b+c-a)(a-c)
$$

Simplification gives $3 b=3 a+c$. Squaring both sides and using $b^{2}=c^{2}+a^{2}$, we obtian $4 c=3 a$. Hence $B C / B A=a / c=4 / 3$.

(If $a \leq c$, then $I$ lies outside the circumcircle of $A E F$.)
