Prove that the altitudes of a triangle are concurrent

1.	Prove that the altitudes of a triangle are concurrent
Answer» Let ABC be any triangle.Let AD ⊥⊥ BC and BE ⊥⊥ toACLet 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 ABLet 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) andb→.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.

Discussion