In ∆ABC, D is the midpoint of BC. If DL ⊥ AB and DM ⊥ AC such that DL=DM, prove that AB=AC.

Question

In &#x2206;ABC, D is the midpoint of BC. If DL &#x22A5; AB and DM &#x22A5; AC such that DL=DM, prove that AB=AC.

Philoid · Accepted Answer

Given: BD = DC and DL ⊥ AB and DM ⊥ AC such that DL=DM

To prove: AB = AC

Proof:

In right angled triangles ∆BLD and ∆CMD,

∠BLD = ∠CMD = 90°

BD = CD … given

DL = DM … given

Thus by right angled hypotenuse side property of congruence,

∆BLD ≅ ∆CMD

Hence, we know that, corresponding parts of the congruent triangles are equal

∠ABD = ∠ACD

In ∆ABC, we have,

∠ABD = ∠ACD

∴ AB= AC …. Sides opposite to equal angles are equal