In a triangle ABC, AB = AC. D is the mid-point of BC, E is the foot of the perpendicular drawn from D to AC, and F is the mid-point of DE. Prove that AF is perpendicular to BE.
I. Definition:
→a=→BC|→a|=a→b=→AC|→b|=b→b⋅→b=b2→a⋅→b=a2⋅a→a⋅→b=a22→d=→AD→d=→b−12→a →f=→AE→f=(→b⋅→d)b2→b→f=(→b⋅(→b−12→a))b2→b→f=(b2−(→a⋅→b)2)b2→b→f=(b2−a24)b2→b→f=(1−a24b2)⋅→b
→e=→ED→e=→d−→f→e=(→b−12→a)−→f→e2=(→b−12→a)12−12→f→AF=→e2+→f=(→b−12→a)12−12→f+→f→AF=(→b−12→a)12+12→f→AF=(→b−12→a)12+12(1−a24b2)⋅→b→AF=→b[12+12⋅(1−a24b2)]−14⋅→a→AF=→b(12+12−a28b2)−14⋅→a→AF=→b(1−a28b2)−14⋅→a →BE=→a−(→b−→f)→BE=→a−→b+→f→BE=→a−→b+(1−a24b2)⋅→b→BE=→a−→b+→b−(a24b2)⋅→b→BE=→a−(a24b2)⋅→b
II. Prove that AF is perpendicular to BE.
→AF=−14⋅→a+→b(1−a28b2)→BE=→a−→b(a24b2)Perpendicular, if ( →AF⋅→BE )=0( →AF⋅→BE )=(−14⋅→a+→b(1−a28b2))⋅(→a−→b(a24b2))( →AF⋅→BE )=−14⋅→a⋅→a+14⋅→a⋅→b(a24b2)+→b⋅→a(1−a28b2)−→b⋅→b(1−a28b2)(a24b2)→a⋅→a=a2→a⋅→b=→b⋅→a=a22→b⋅→b=b2( →AF⋅→BE )=−a24+14⋅a22(a24b2)+a22(1−a28b2)−b2(1−a28b2)(a24b2)( →AF⋅→BE )=−a24+a432b2+a22−a416b2−(1−a28b2)(a24)( →AF⋅→BE )=−a24+a432b2+a22−a416b2−a24+a432b2( →AF⋅→BE )=−a24−a24+a22+a432b2+a432b2−a416b2( →AF⋅→BE )=−a22+a22+a416b2−a416b2=0
