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Final Round of National Olympiad

Geometry Difficulty 4.9 AIME Prove it Estonia

The bases of trapezoid ABCDABCD are ABAB and CDCD, and the intersection point of its diagonals is PP. Prove that if ∣PA∣∣PD∣=∣PB∣∣PC∣\frac{|PA|}{|PD|} = \frac{|PB|}{|PC|} then the trapezoid is isosceles.

Solutions — 2

Solution 1

Figure 1
Figure 15

By assumptions, ∣PA∣∣PB∣=∣PD∣∣PC∣\frac{|PA|}{|PB|} = \frac{|PD|}{|PC|}. As the bases ABAB and CDCD are parallel, we have also ∣PA∣∣PB∣=∣PC∣∣PD∣\frac{|PA|}{|PB|} = \frac{|PC|}{|PD|} (Fig. 15). Hence ∣PC∣=∣PD∣|PC| = |PD|. Similarity of triangles APDAPD and BPCBPC implies ∣AD∣∣BC∣=∣PD∣∣PC∣=1\frac{|AD|}{|BC|} = \frac{|PD|}{|PC|} = 1, thus ∣AD∣=∣BC∣|AD| = |BC| as needed.

Solution 2

Figure 2
Figure 16

Figure 3
Figure 17

By assumptions, triangles APDAPD and BPCBPC are similar. Thus ∠ADB=∠ACB\angle ADB = \angle ACB (Fig. 16), showing that quadrilateral ABCDABCD is cyclic. But if a quadrilateral with parallel opposite sides has a circumcircle, the bisectors of these sides coincide as they have the same direction and both pass through the circumcenter of the quadrilateral (Fig. 17). By symmetry w.r.t. this line, the other pair of opposite sides have equal lengths.

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