If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is

It is given in the question that,

In ∆ABC, BL is parallel to AC

Also, CM is parallel AB such that BL = CM

We have to prove that: AB = AC

Now, in ∆ABL and ∆ACM we have:

BL = CM (Given)

∠BAL = ∠CAM (Common)

∠ALB = ∠AMC (Each angle equal to 90^o)

∴ By AAS congruence criterion

∆ABL ≅ ∆ACM

AB = AC (By Congruent parts of congruent triangles)

As opposite sides of the triangle are equal, so it is an isosceles triangle

Hence, option (B) is correct