Q. 134.5( 2 Votes )

# If the altitudes

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

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