prove that altitudes of a triangle r concurrent

Dear keshav kr,

Now this is how you can  prove the concurrentcy of the altitudes of a triangle .

Given triangle ABC with altitudes: AE, BD and CF we want to show that they meet in one point.

Through each of the vertices of the triangle construct a line parallel to the opposite side of the triangle forming triangle PQR.

Now RA = BC since RACB is a parallelogram,
Also AQ = BC since ABCQ is a parallelogram,
Hence RA = AQ 

AE is perpendicular to RQ since AE is perpendicular to BC and BC is parallel to RQ

Hence AE is the perpendicular bisector of RQ.

Similarly BD is the perpendicular bisector of RP and CF the perpendicular bisector of QP.

AE, BD and CF are therefore concurrent since we know that the perpendicular bisectors of the sides of a triangle are concurrent at a point called the circumcentre(C).

Hope this helped you immensely..

All the Very Best & Good Luck to you ...

Regards,

AskIITians Expert,

Godfrey Classic Prince

IIT-Madras

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