Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
Click the figure below to view more details of problem 1206.
Saturday, April 9, 2016
Geometry Problem 1206: Circle, Angle Bisector, Secant, Triangle, Circumcircle, Congruence, Area
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Lets assume angle AMB = 2x
ReplyDeleteAngle PMR = x and angle PMB = 180 - x. Hence arc PR = arc PB and segment PR = PB
Similarly PQ = PA
Angle RMB = Angle AMQ, hence Angle RPB = Angle QPA = y
Triangle APR is congruent to triangle QPB, by SAS, hence AR=BQ
Area of triangle PRC = PR.PC.sin (y)= PB.PC.sin(y)
Area of triangle PQD = PQ.PD.sin(y) = PA.PD.sin(y)
Since PA.PD = PB.PC, area of the triangles are equal.
Well done
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