Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 140Triangle, Excircle, Tangent, Semiperimeter. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
A circle O tangent to BC at F.We have :AD=AE , BD=BF & CE=CF.p=(AB+AC+BC)/2p=(AB+AC+BF+CF)/2p=(AB+AC+BD+CE)/2p=(AD+AE)/2p=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:angle AEO and angle ADO are right angles because a tangent is perpendicular to a radiusAO = AO by reflexive propertyOD = OE =rso triangle AOD is congruent to triangle AEOby hypotenuse leg.and AD = AE by CPCTC
To Dan May (about problem 140).The 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!