Difference between revisions of "1971 Canadian MO Problems/Problem 3"
Airplanes1 (talk | contribs) (Created page with "== Problem == <math>ABCD</math> is a quadrilateral with <math>AD=BC</math>. If <math>\angle ADC</math> is greater than <math>\angle BCD</math>, prove that <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>. | 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>. | ||
| + | == 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]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 15:48, 12 September 2012
Problem
is a quadrilateral with 
. If 
is greater than 
, prove that 
.
Solution
Consider triangles 
and 
. These triangles share two of the same side lengths. Thus, by the Hinge Theorem, since we are given that 
, we have .
See Also
| 1971 Canadian MO (Problems) | ||
| Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 4 |
