Unexpected proof of Base angles of isosceles triangle theorem
The base angles of isosceles triangle are equal. This fact is proved by either one of the following methods in most geometry books:
(1) Let M be the mid point of BC. Join AM. AB = AC (given) BM = MC (construction) AM = AM (common) \ DABM º DACM (SSS) \ ÐABC = ÐACB (corr. Ðs of º Ds)
(2) From A construct a line AM such that: AM ^ BC. AB = AC (given) ÐAMB = ÐAMC = 90° (construction) AM = AM (common) \ DABM º DACM (RHS) \ ÐABC = ÐACB (corr. Ðs of º Ds)
The proofs involve the construction of a straight line. Here is an interesting method without the need to construct any straight line. The proof is in fact generated by computer program!
AB = AC (given) AC = AB (given) ÐBAC = ÐCAB (common) \ DABC º DACB (SAS) \ ÐABC = ÐACB (corr. Ðs of º Ds) Bingo!