Question

AD and BC are equal perpendiculars drawn on AB. Prove that DC bisects AB.

Answer 1

Given:

AD ⊥ AB and BC ⊥ AB, where AD = BC.

We need to prove that DC bisects AB, i.e., AC = CB.

Proof:

1. Draw the figure:

Let A, B and C be three points.

Let D be a point on line AB, such that AD ⊥ AB and BC ⊥ AB, where AD = BC.

2. Using symmetry:

Since AD = BC and both are perpendicular to AB, triangles △ABD and △CBA are right-angled triangles by the property of perpendicular lines.

3. Congruence of triangles:

We now have two right-angled triangles: △ABD and △CBA.

In these triangles:

AD = BC  ...(Given)

AB is a common side,

∠ABD = ∠CBA = 90° since both are right angles.

By Hypotenuse-Leg (HL) congruence, these two triangles are congruent △ABD ≅ △CBA

4. Conclusion from congruence:

From the congruence of triangles △ABD and △CBA, we can conclude that the corresponding sides are equal AC = BC.

This implies that DC bisects AB because it divides AB into two equal parts.

Thus, DC bisects AB.