2008 AMC 10B Problems/Problem 24
Problem
Quadrilateral has , angle and angle . What is the measure of angle ?
Solution
Solution 1
Draw the angle bisectors of the angles and . These two bisectors obviously intersect. Let their intersection be . We will now prove that lies on the segment .
Note that the triangles and are equal, as they share the side , and we have and .
Also note that for similar reasons the triangles and are equal.
Now we can compute their inner angles. is the bisector of the angle , hence , and thus also . is the bisector of the angle , hence , and thus also .
It follows that . Thus the angle has , and hence does indeed lie on . Then obviously .
Solution 2
Draw the diagonals and , and suppose that they intersect at . Then, and are both isosceles, so by angle-chasing, we find that , CBD = 5^{\circ}\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}E'EE'B = 60^{\circ}E'\overline{AE}E\angle EEC = 60^{\circ}E\overline{DE}\triangle BEE'\triangle CEE\triangle BEC \cong \triangle DEC\triangle BCE \cong \triangle BAE'DE = BE = EE'EE = CE = E'AAE = AE' + E'E = EE + DE = DE\triangle ADE\angle AED = 120^{\circ}\angle DAC = \frac{180 - 120}{2} = 30^{\circ}\angle BAD = 30 + 55 = 85^{\circ}$.
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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