### 2006 IMO #G3

Let $ABCDE$ be a convex pentagon such that $\angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.$The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Solution 1:

Note that $\triangle ABC \sim \triangle ADE \sim \triangle ACD$ from AA, so $\frac{BC}{DC}=\frac{AC}{AD}=\frac{CD}{DE}$ as well as $\angle BCD = \angle CDE$, so $\triangle CBD \sim \triangle DCE$. Let $DB \cap AC = Q, EC \cap AD = R$. Then from Ceva's Theorem, it suffices to show that $\triangle ARQ \sim \triangle ADC$. We also have $\angle CBM = \angle DCE$, so we get $\triangle DRC \sim \triangle CQB$. This means that $\frac{BC}{DC}=\frac{QC}{RD}=\frac{AC}{AD}$, so $\triangle ARQ \sim \triangle ADC$, as desired. $\square$