Balti Tee 1990 · Ülesanne 6

Geomeetria

Let $A B C D$ be a quadrangle, $|A D|=|B C|, \angle A+\angle B=120^{\circ}$ and let $P$ be a point exterior to the quadrangle such that $P$ and $A$ lie at opposite sides of the line $D C$ and the triangle $D P C$ is equilateral. Prove that the triangle $A P B$ is also equilateral.

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Solution:

Note that $\angle ADC + \angle CDP + \angle BCD + \angle DCP = 360^{\circ}$ (see Figure 1). Thus $\angle ADP = 360^{\circ} - \angle BCD - \angle DCP = \angle BCP$ . As we have $|DP| = |CP|$ and $|AD| = |BC|$ , the triangles $ADP$ and $BCP$ are congruent and $|AP| = |BP|$ . Moreover, $\angle APB = 60^{\circ}$ since $\angle DPC = 60^{\circ}$ and $\angle DPA = \angle CPB$ .

Diagram for the mathnet 00wh 1 of bw-1990-06. Figure 1

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AoPS · Baltic Way 1990