Difference between revisions of "2019 AMC 10A Problems/Problem 13"
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==Solution== | ==Solution== | ||
| + | Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Noting that they subtend to the same arc of <math>180^{\circ}</math>, we can find <math>\angle ECB=20</math> and <math>\angle DBC=50</math> by the triangle angle sum. Then we take triangle <math>BFC</math>, and find <math>\angle BFC=180-50-20=\boxed{(\text{D})110}.</math> | ||
==See Also== | ==See Also== | ||
Revision as of 18:02, 9 February 2019
Problem
Let 
be an isosceles triangle with 
and 
. Contruct the circle with diameter 
, and let 
and 
be the other intersection points of the circle with the sides 
and 
, respectively. Let 
be the intersection of the diagonals of the quadrilateral 
. What is the degree measure of 
Solution
Drawing it out, we see 
and 
are right angles, as they are inscribed in a semicircle. Noting that they subtend to the same arc of 
, we can find 
and 
by the triangle angle sum. Then we take triangle 
, and find 
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
