Problem 6

Consider an isosceles triangle. Let

R

be the radius of its circumcircle and

r

the radius of its inscribed circle. Prove that the distance

d

between the centers of these two circles is

\sqrt{R(R-2r)}

Step 3 of 6: Show that the auxiliary triangle is isosceles

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\angle BIL=\angle IBL=\frac12\angle A+\frac12\angle B\Longrightarrow IL=BL

Detailed analysis

Because $BI$ bisects angle $B$ and the cyclic angles subtending the same arcs have the required equal values, the source angle chase gives $\angle BIL=\angle IBL=\frac12\angle A+\frac12\angle B$ . Therefore $\triangle BIL$ is isosceles and $IL=BL$ .