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

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

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## Thursday, January 14, 2016

### Geometry Problem 1178: Triangle, Internal Angle Bisector, Metric Relations

Labels:
angle bisector,
metric relations,
triangle

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Lets assume AD extended meets circumcircle of Tr.ABC at point E.

ReplyDeleteAssuming DE=x, we have d.x= BD.CD

BD = ac/(b+c), CD = ab/(b+c)

Also Tr. ACD is similar to Tr.AEB

We have d/c=b/(d+x) we get

d^2 = bc - dx or d^2 = bc - BD.CD

Substituting values of BD and CD we get required expression.

Perhaps the easiest way of doing this thro' pure geometry alone

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