定理:任意三角形都是等腰三角形.这显然是荒谬的命题!怎么会正确呢?但有人恰能证明如下:已知:△ABC中,AB≠AC.求证:AB=AC.证明:设△ABC中∠A的平分线与边BC的中垂线相交于点O,连结AO、BO、CO.过点O作ON⊥AB,OK⊥AC,N,K是垂足.∵AO平分∠A,∴ON=OK,AN=AK.∵MO垂直平分BC,∴BO=CO Theorem: Any triangle is an isosceles triangle. This is obviously a ridiculous proposition! How can it be correct? But some people just can prove as follows: Known: △ ABC, AB ≠ AC. Proof: AB = AC. Proof: Set △ The bisector of A in ABC intersects with the mid-perpendicular of edge BC at point O, linking AO, BO, and CO. The over-point O acts as ON⊥AB, OK⊥AC, N, and K are the perpendicular feet. , ∴ON=OK, AN=AK.∵MO Vertically divide BC, ∴BO=CO