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