Prove that the altitudes of a …

Let ABC be any triangle. Let AD ⊥⊥ BC and BE ⊥⊥ toAC Let AD and BE intersect at O (origin say) Join CO and extend it to meet AB at F ⇒⇒ AOD and BOE are two altotudes of the triangle. We have to prove that COF is the third altitude. ⇒⇒ we have to prove that CF is ⊥⊥ to AB Let OA−→−=a→,OB−→−=b→OA→=a→,OB→=b→ and OC−→−=c→OC→=c→ We know that AB−→−=b→−a→,BC−→−=c→−b→andAC−→−=c→−a→AB→=b→−a→,BC→=c→−b→andAC→=c→−a→ Since AD ⊥⊥ BC and BE ⊥⊥ AC, a→.(c→−b→)=0andb→.(c→−a→)=0a→.(c→−b→)=0andb→.(c→−a→)=0 ⇒a→.c→=a→.b→⇒a→.c→=a→.b→..........(i) and b→.c→=b→.a→b→.c→=b→.a→..........(ii) But we know that a→.b→=b→.a→a→.b→=b→.a→ ⇒(i)=(ii)⇒(i)=(ii) ⇒a→.c→=b→.c→⇒a→.c→=b→.c→ ⇒a→.c→−b→.c→=0⇒a→.c→−b→.c→=0 ⇒(a→−b→).c→=0⇒(a→−b→).c→=0 ⇒a→−b→is⊥toc→⇒a→−b→is⊥toc→ ⇒AB−→−⊥toOC−→−⇒AB→⊥toOC→ ⇒⇒ FOC is altitude of the side AB ⇒⇒ All the three altitudes meet at a common point O.