2006 Romanian NMO Problems/Grade 9/Problem 2


Let $ABC$ and $DBC$ be isosceles triangles with the base $BC$. We know that $\angle ABD = \frac{\pi}{2}$. Let $M$ be the midpoint of $BC$. The points $E,F,P$ are chosen such that $E \in (AB)$, $P \in (MC)$, $C \in (AF)$, and $\angle BDE = \angle ADP = \angle CDF$. Prove that $P$ is the midpoint of $EF$ and $DP \perp EF$.


Since $\triangle BDC$ is isosceles, $BD=DC$. Since $\angle ABD = \dfrac{\pi}{2}$, $\dfrac{\pi}{2} = \angle ABD = \angle ABC + \angle DBC = \angle ACB + \angle DCB = \angle ACD$, which means that $\angle DCF = \dfrac{\pi}{2}$, too. Thus $\angle EBD = \angle DCF$, so by ASA, $\triangle EBD \cong \triangle FCD$. This means that $ED = FD$. Since $AB = AC$, $BD = DC$, and $\angle ABD = \angle ACD$, by SAS, $\triangle ABD \cong \triangle ACD$, so $\angle BDA = \angle ADC$. Since $\angle BDE = \angle MDP$, $\angle EDA = \angle BDA - \angle BDE = \angle ADC - \angle MDP = \angle CDP$. Thus $\angle EDP = \angle EDA + \angle MDP = \angle PDC + \angle CDF = \angle PDF$. $DP$ is the angle bisector of $\angle EDF$, and $ED = DF$. This means that $EF \perp DP$. $AM = AM$, $BM = MC$, and $AB = AC$, so by SSS, $\triangle ABM \cong \triangle ACM$. Thus $\angle AMC = \dfrac{\pi}{2}$ and $\angle DMP = \dfrac{\pi}{2}$. $\angle MDP = \angle BDE$, so by AA, $\triangle BDE \sim \triangle MDP$. Thus $\dfrac{ED}{BD} = \dfrac{DP}{MD}$. Also, $\angle EDF = \angle EDC + \angle CDF = \angle EDC + \angle EDB = \angle BDC$. $BD = DC$ and $ED = DF$, so $\dfrac{BD}{ED} = \dfrac{DC}{DF}$. By SAS similarity, $\triangle BDC \sim \triangle EDF$. $MD$ is a median and an angle bisector of $\triangle BDC$. Now assume that P' is the point such that DP' is a median of $\triangle EDF$ (it is on $EF$). It is on DP, the angle bisector, and since $\triangle BDC \sim \triangle EDF$, $\dfrac{ED}{BD} = \dfrac{DP'}{MD}$, but we also showed that $\dfrac{ED}{BD} = \dfrac{DP}{MD}$. Thus $DP' = DP$. Since P and P' are on the same ray ($DP$), P = P' and P is the midpoint of $EF$.

See also