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
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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!