Difference between revisions of "1990 AHSME Problems/Problem 20"
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− | Let <math>\angle DAE = \alpha</math> and <math>\angle | + | Let <math>\angle DAE = \alpha</math> and <math>\angle ADE = \beta</math>. |
Since <math>\alpha + \beta = 90^\circ</math>, and <math>\alpha + \angle BAF = 90^\circ</math>, then <math>\beta = \angle BAF</math>. By the same logic, <math>\angle ABF = \alpha</math>. | Since <math>\alpha + \beta = 90^\circ</math>, and <math>\alpha + \angle BAF = 90^\circ</math>, then <math>\beta = \angle BAF</math>. By the same logic, <math>\angle ABF = \alpha</math>. |
Latest revision as of 17:05, 23 March 2020
Problem
In the figure is a quadrilateral with right angles at and . Points and are on , and and are perpendicual to . If and , then
Solution
Label the angles as shown in the diagram. Since forms a linear pair with , is a right angle.
Let and .
Since , and , then . By the same logic, .
As a result, . By the same logic, .
Then, , and .
Then, , and .
By the transitive property, . , and plugging in, we get .
Finally, plugging in to , we get
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.