Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Wednesday, January 20, 2016

### Geometry Problem 1181: Cyclic Quadrilateral and Tangential Quadrilateral, Diameter as a Diagonal, Incenter, Circumcenter

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Ang B1=Ang B2 (BO bisector), => D1=D2 => AB=BC

ReplyDeleteBut BC-BF=3 => x=3

BD is a diameter of circle ABCD and also bisects < ABC. Hence Tr. s ABD and BCD are congruent ASA.

ReplyDeleteSo AB = BC

Hence x = BC - BF

Now BC + 7 = BF + 10 since BCEF is a tangential quadrilateral

So BC - BF = 3 and hence x = 3

Sumith Peiris

Moratuwa

Sri Lanka

Triangles BCD and BAD are right triangles

ReplyDelete∠ (CBD)= ∠ (ABD) => Tri BCD congruence to BAD and BA=BC

BCEF is a tangential quadrilateral so

BF+CE=EF+BC => BC-BF=CE-EF= 10-7=3

BA-BF=FA=3