Difference between revisions of "1971 Canadian MO Problems/Problem 3"

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== Solution ==
 
== Solution ==
  
Consider triangles <math>ADC</math> and <math>BDC</math>. These triangles share two of the same side lengths. Thus, by the Hinge Theorem, since we are given that <math>\angle ADC    \ge    \angle BCD</math>, we have <math>AC>BD</math>.
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Consider triangles <math>ADC</math> and <math>BDC</math>. These triangles share two of the same side lengths. Thus, by the [[Hinge Theorem]], since we are given that <math>\angle ADC    \ge    \angle BCD</math>, we have <math>AC>BD</math>.
  
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== See Also ==
 
{{Old CanadaMO box|num-b=2|num-a=4|year=1971}}
 
{{Old CanadaMO box|num-b=2|num-a=4|year=1971}}
[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Geometry Problems]]

Latest revision as of 12:29, 10 December 2019

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