Polish MO finals, problem 3

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Polish MO finals, problem 3

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#1 h Apr 10, 2008, 6:41 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

In a convex pentagon $ABCDE$ in which $BC=DE$ following equalities hold:

$\angle ABE =\angle CAB =\angle AED-90^{\circ},\qquad \angle ACB=\angle ADE$

Show that $BCDE$ is a parallelogram.

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#2 h Apr 10, 2008, 10:53 PM • 4 Y

Y by Adventure10, Mango247, Rijul saini, and 1 other user

Let $EB \cup DA = G$ and $EB \cup CA = F$ , and $\angle CAB = \angle EBA = \phi, \angle ADE = \angle ACB = \varphi$ . Then, by sine law to triangles $DEA$ , $EAB$ and $ABC$ with $DE = CB$ in mind, we get that $\angle AEB = \angle GAE$ ( $\frac {DE}{EA} \frac{EA}{AB} \frac{AB}{CB}=\frac{\sin \angle EAG}{\sin \varphi} \frac{\sin \phi}{\sin \angle EAB} \frac {\sin \varphi} {\sin \phi}$ ).
Let $X$ and $Y$ be intersectio point of perpendicular line from $E$ to $DE$ with $DA$ and $CA$ respectively. Then, $\angle XEA = \phi$ , so $\angle EXD = \phi + \angle EAG$ . $\angle BFY = 2\phi$ so $\angle FYE = \phi + \angle GEA = \phi + \angle EAG = \angle EXD$ so $EX\bot CB$ , hence $CB|| DE$ .

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