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

Labels:
angle bisector,
area,
circle,
circumcircle,
congruence,
secant,
triangle

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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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