Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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We will now prove that <math>P</math> lies on the segment <math>AD</math>. | We will now prove that <math>P</math> lies on the segment <math>AD</math>. | ||
− | Note that the triangles <math>ABP</math> and <math>CBP</math> are | + | Note that the triangles <math>ABP</math> and <math>CBP</math> are congruent, as they share the side <math>BP</math>, and we have <math>AB=BC</math> and <math>\angle ABP = \angle CBP</math>. |
− | Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are | + | Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent. |
Now we can compute their inner angles. <math>BP</math> is the bisector of the angle <math>ABC</math>, hence <math>\angle ABP = \angle CBP = 35^\circ</math>, and thus also <math>\angle CDP = 35^\circ</math>. <math>CP</math> is the bisector of the angle <math>BCD</math>, hence <math>\angle BCP = \angle DCP = 85^\circ</math>, and thus also <math>\angle BAP = 85^\circ</math>. | Now we can compute their inner angles. <math>BP</math> is the bisector of the angle <math>ABC</math>, hence <math>\angle ABP = \angle CBP = 35^\circ</math>, and thus also <math>\angle CDP = 35^\circ</math>. <math>CP</math> is the bisector of the angle <math>BCD</math>, hence <math>\angle BCP = \angle DCP = 85^\circ</math>, and thus also <math>\angle BAP = 85^\circ</math>. |
Revision as of 17:36, 1 January 2013
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 congruent, as they share the side , and we have and .
Also note that for similar reasons the triangles and are congruent.
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 , , and . Draw such that and so that is on , and draw such that and is on . It follows that and are both equilateral. Also, it is easy to see that and by construction, so that and . Thus, , so is isosceles. Since , then , and .
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |