Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Estonia

The bases of trapezoid $ABCD$ are $AB$ and $CD$ , and the intersection point of its diagonals is $P$ . Prove that if $\frac{|PA|}{|PD|} = \frac{|PB|}{|PC|}$ then the trapezoid is isosceles.

Solutions — 2

Solution 1

Figure 15

By assumptions, $\frac{|PA|}{|PB|} = \frac{|PD|}{|PC|}$ . As the bases $AB$ and $CD$ are parallel, we have also $\frac{|PA|}{|PB|} = \frac{|PC|}{|PD|}$ (Fig. 15). Hence $|PC| = |PD|$ . Similarity of triangles $APD$ and $BPC$ implies $\frac{|AD|}{|BC|} = \frac{|PD|}{|PC|} = 1$ , thus $|AD| = |BC|$ as needed.

Solution 2

Figure 16

Figure 17

By assumptions, triangles $APD$ and $BPC$ are similar. Thus $\angle ADB = \angle ACB$ (Fig. 16), showing that quadrilateral $ABCD$ 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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