IIT JEE 1989 let abc be a triangle with ab=ac. If D is the midpoint of BC, E is the foot of the perpendicular drawn from D to AC and F the midpoint of DE, Prove that AF is perpendicular to BE.

This is a coordinate geometry problem in which the key is to set up the triangle with an appropriately chosen coordinate system. So you align the triangle so that it sits on the x-axis with the origin as the mid point of the base. The  points B and C defining the base are (b, 0) and (-b, 0). The vertex (0, a) lies on the Y-axis,  since the triangle is isosceles and median coincides with altitude. Now all we have to so is to find the vector directions of AF and BE in terms of a and b, and show that their dot product is 0. The coordinates of E  can be worked out by using the fact that the triangles ADC and DEC are similar.   Once you have the coordinates of E, you can also work out the coordinates of F as the mid point of D and E. This provides you all the information you need to work out the vector directions AF and BE, because the coordinates of all end points are known. Now compute the dot product in  terms of a and b and show that the algebraic expression computes to 0.

 

Incidentally, I actually faced this problem in the real IIT exam in 1989, and the problem is very memorable to me for a particular reason :)

 

Charu Aggarwal