See complete Problem 140

Triangle, Excircle, Tangent, Semiperimeter. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Thursday, July 24, 2008

### Elearn Geometry Problem 140

Labels:
circle,
congruence,
excircle,
semiperimeter,
tangent,
triangle

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A circle O tangent to BC at F.

ReplyDeleteWe have :AD=AE , BD=BF & CE=CF.

p=(AB+AC+BC)/2

p=(AB+AC+BF+CF)/2

p=(AB+AC+BD+CE)/2

p=(AD+AE)/2

p=AD=AE

When you say CE=CF and BD=BF you are using the same principle you are trying to prove. That a point is equal distant from the two tangent points. How about:

ReplyDeleteangle AEO and angle ADO are right angles because a tangent is perpendicular to a radius

AO = AO by reflexive property

OD = OE =r

so triangle AOD is congruent to triangle AEO

by hypotenuse leg.

and AD = AE by CPCTC

To Dan May (about problem 140).

ReplyDeleteThe goal of the problem is not to prove that AD = AE. It is to prove that both are equal to the semiperimeter p of the triangle ABC!