Difference between revisions of "1971 Canadian MO Problems/Problem 3"
(→See Also) |
m (→Solution) |
||
| Line 4: | Line 4: | ||
== 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>. | + | 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 == | == 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 Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Latest revision as of 11:29, 10 December 2019
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 |
