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

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

Click the figure below to see the complete problem 874.

## Monday, May 13, 2013

### Problem 874: Isosceles Triangle, Circumcircle, Circumcenter, Congruence

Labels:
circle,
circumcenter,
circumcircle,
congruence,
isosceles,
triangle

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Let the mid point of O1O2 be M, M must lies on BD (the radical axis of two circles)

ReplyDeleteAngle(BO1O2)

=angle(BO1M)

=angle(BAD)

=angle(BCD)

=angle(BO2M)

=angle(BO2O1)

So BO1O2 forms an isosceles triangle

Q.E.D.

build:

ReplyDeleteO2D

O2B

O1B

O1D

A=C=a

BO2D=2a=BO1D

O2BD=O2DB=O1BD=O1DB=90-a

O1BO2=O1DO2=180-2a

BO1DO2 parallelogram

O2D=O1B

circumcircle O1 and O2 are congruent

Q.E.D.

2×BO1 = BD / sin∠BAC = BD / sin∠BCA = 2×BO2

ReplyDeleteThe result follows.

< BO1D = 2A hence < ABO1 = 90-A - < ABD

ReplyDeleteSimilarly < CBO2,= 180 - 2A - < ABD - (90-A) = < ABO1

So Tr.s ABO1 and BCO2 are congruent ASA and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

It also follows that BO1DO2 is a rhombus so BD and O1O2 bisect each other perpendicularly

ReplyDelete