Question

Abc is an isosceles triangle with ab=ac.D is the mid point of bc.Dm perpendicular on ab and dn perpendicular on ac.

Answer 1

Here is a solution that doesn’t require congruency. We denote area of △XYZ△XYZby [△XYZ],[△XYZ], and the distance from a point PP to a line ℓℓ by dist(P,ℓ)dist(P,ℓ)

Then we know that [△ABD]=[△ACD],[△ABD]=[△ACD], because each one is equal to 14⋅BC⋅dist(A,BC).14⋅BC⋅dist(A,BC).

Now we compute the areas in a different manner.

12⋅AC⋅dist(D,AC)=[△ACD]=[△ABD]=12⋅AB⋅dist(D,AB)=12⋅AB⋅dist(D,AC)12⋅AC⋅dist(D,AC)=[△ACD]=[△ABD]=12⋅AB⋅dist(D,AB)=12⋅AB⋅dist(D,AC)

So we finally have AB=ACAB=AC, which was required ■