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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.1 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle with $\angle B \leq \angle C$ , $I$ its incenter and $D$ the intersection point of line $AI$ with side $BC$ . Let $M$ and $N$ be points on sides $BA$ and $CA$ , respectively, such that $BM = BD$ and $CN = CD$ . The circumcircle of triangle $CMN$ intersects again line $BC$ at $P$ . Prove that quadrilateral $DIM P$ is cyclic.

Solution

By the bisector theorem we have
$\frac{MB}{AB} = \frac{DB}{AB} = \frac{DC}{AC} = \frac{NC}{AC}$
We deduce that segments $MN$ and $BC$ are parallel.

Because $CNMP$ is a cyclic trapezoid, it is isosceles. Hence
$\angle DPM = \angle ACB$
Because $BD = BM$ and $I$ is on the bisector of $\angle DBM$ , the quadrilateral $BDIM$ is a kite. We deduce that
$\begin{aligned} \angle MID & = 360^\circ - \angle DBM - 2 \angle ADB \\ & = 360^\circ - \angle CBA - 2\left(\angle ACB + \frac{1}{2} \angle BAC\right) \\ & = 180^\circ - \angle ACB = 180^\circ - \angle DPM. \end{aligned}$
This proves that quadrilateral $DIM P$ is cyclic.

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