in triangleABC D is the mid point of BC if DLperpendicular AB and DM perpendicuarAC such that DL=DM prove that AB=AC
To prove that AB = AC in triangle ABC, given that D is the midpoint of BC and DL is perpendicular to AB while DM is perpendicular to AC with DL = DM, we can use the concept of congruent triangles.
Here's how we can proceed with the proof:
1. Draw the given triangle ABC with D as the midpoint of BC and DL perpendicular to AB, while DM is perpendicular to AC.
2. Since D is the midpoint of BC, we can conclude that AD is a median of triangle ABC. This means AD divides the opposite side BC into two equal segments.
3. Since DL is perpendicular to AB and DM is perpendicular to AC, we have right triangles ABD and ACD.
4. As DL = DM, the hypotenuses of the right triangles ABD and ACD are equal.
5. Now, we need to prove that the corresponding sides of the right triangles ABD and ACD are equal.
6. In right triangle ABD, we can say that side AD is common to both triangles, and side DL is equal to side DM.
7. By the hypotenuse-leg congruence criterion, triangles ABD and ACD are congruent, which means the corresponding sides are equal.
8. Hence, we can conclude that AB = AC, as all the corresponding sides of triangles ABD and ACD are equal.
Therefore, we have proved that in triangle ABC, if D is the midpoint of BC, DL is perpendicular to AB, and DM is perpendicular to AC with DL = DM, then AB = AC.