1971 Canadian MO Problems/Problem 3

Problem

$ABCD$ is a quadrilateral with $AD=BC$ . If $\angle ADC$ is greater than $\angle BCD$ , prove that $AC>BD$ .

Solution

Consider triangles $ADC$ and $BDC$ . These triangles share two of the same side lengths. Thus, by the Hinge Theorem, since we are given that $\angle ADC \ge \angle BCD$ , we have $AC>BD$ .

1971 Canadian MO (Problems)
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 4

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1971 Canadian MO Problems/Problem 3

Problem

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