In triangle ABC, altitudes are drawn from the vertices B and C on AC and AB respectively such that BL=CM. Prove that Triangle BCM congruent triangle CBL.

Kindly give answer to the above question?

**Given that:**

In ΔABC, altitude are drawn from the vertices B and C on AC and AB respectively such that BL = CM.

**To prove:** ΔBCM ΔCBL

Since BL and CM are altitudes

⇒ ∠ALB = ∠AMC = 90°

So, ΔBCM and ΔCBL are right angled triangle.

Also given BL = CM and BC is a common side in both triangles.

⇒ Hypotenuse and one side of triangle BCM is equal to hypotenuse and one side of other side right angled triangle.

Hence by RHS congruence criterion, ΔBCM ΔCBL

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