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

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

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## Thursday, December 6, 2018

### Geometry Problem 1405: Four Congruent Circles, Congruent Chords

Labels:
chord,
congruence,
four circles,
geometry problem,
intersecting circles

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From ang CFH and CFP => CH = CP

ReplyDeleteFrom congr of circles AC = BC

=> Tr ACH congr Tr BCP (SAS)

ABC, BCD, BDF are all equilateral triangles

ReplyDeleteABF is a straight line and a diameter of circle B and so AH is perpendicular to FP

S(BFP) = 1/2.BP/2.FP......(1) (since < BPF = 30 since D is the centre of TR BFP)

S(AFP) = 1/2.AH. FP ......(2)

Since S(AFP) = 2.S(BFP) from (1) and (2) we have

AH = BP

Sumith Peiris

Moratuwa

Sri Lanka

CBF equal 120*. Angle CPF equal 60*. Angle CHP=FAC=CPE=60*.Triangle CHP equilateral. PC=PH=HC. R(C,60*) B-->A P-->H BP=AH. In triangle APH AP in second power=AH second power+ HC in second power=BP in second power + PC in second power. This means AP in second power=Bp in second power + PC in second power.

ReplyDeleteErina,NJ