Saturday, April 9, 2016

Geometry Problem 1206: Circle, Angle Bisector, Secant, Triangle, Circumcircle, Congruence, Area

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.

Geometry Problem 1206: Circle, Angle Bisector, Secant, Triangle, Circumcircle, Congruence, Area
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2 comments:

  1. Pradyumna AgasheApril 10, 2016 at 3:16 AM

    Lets assume angle AMB = 2x
    Angle 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.

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