1971 Canadian MO Problems/Problem 3

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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 $AC>BD$.

See Also

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