Question

In ΔABC, AB = AC and CP and BQ are altitudes. Prove that CP = BQ.

Answer 1

Given:

ΔABC with AB = AC Isosceles triangle with vertex A and base BC.

CP and BQ are altitudes.

To Prove: CP = BQ.

Step 1: Recall properties of altitudes in isosceles triangles

In an isosceles triangle AB = AC.

Altitudes drawn from the base vertices B and C to the opposite sides here, from C to AB and from B to AC are equal in length.

We need to prove it formally.

Step 2: Consider right triangles formed by the altitudes 

Let P be the foot of the perpendicular from C to AB.

Let Q be the foot of the perpendicular from B to AC.

Then we have right triangles:

1. ΔCPB right triangle at P.
2. ΔBQA right triangle at Q.

Step 3: Show congruence of the right triangles

Consider ΔCPB and ΔBQA:

Sides: AB = AC  ...(Given)

Right angles: ∠CPB = ∠BQA = 90°

Hypotenuse: CB = CB  ...(Common side)

By RHS congruence criterion Right angle, Hypotenuse, Side:

ΔCPB ≅ ΔBQA

Step 4: Conclude the altitudes are equal

Corresponding sides of congruent triangles are equal CP = BQ.