Difference between revisions of "2019 AMC 10A Problems/Problem 13"
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~Argonauts16 (Diagram by Brendanb4321) | ~Argonauts16 (Diagram by Brendanb4321) | ||
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+ | Alternatively, we could have used similar triangles. | ||
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+ | Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Therefore, <math>\angle ADB = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}</math>. So, <math>\triangle BEF \sim BDA</math> by AA Similarity, since <math>\angle EBF = \angle DBA</math>. Thus, we know <math>\angle EFB = \angle DAB = \angle CAB = 70^{\circ}</math>. Finally, we know that <math>\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.</math> | ||
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+ | ~ alleycat | ||
==Solution 2== | ==Solution 2== |
Revision as of 12:34, 11 February 2019
Contents
Problem
Let be an isosceles triangle with and . Construct 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 1
Drawing it out, we see and are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find . We can find and by the triangle angle sum on and .
Then, we take triangle , and find
~Argonauts16 (Diagram by Brendanb4321)
Alternatively, we could have used similar triangles.
Drawing it out, we see and are right angles, as they are inscribed in a semicircle. Therefore, . So, by AA Similarity, since . Thus, we know . Finally, we know that
~ alleycat
Solution 2
Through the property of angles formed by intersecting chords, we find that
Through the Outside Angles Theorem, we find that
Adding the two equations gives us
Since is the diameter, and because is isosceles and , . Thus
~mn28407
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.