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 90o)
∴ 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