Question

In Fig, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP ∥ AC and RP ∥ AB. Find ∠P.

Answer 1

In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.

∠QCA = ∠CQP (Alternate interior angles)

Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,

∠ABC = ∠PRQ (alternate interior angles).

We know that the sum of all three angles of a triangle is 180°.

Hence, for ∆ABC, we can say that:

∠ABC + ∠ACB + ∠BAC = 180°

∠ABC + ∠ACB + 90° = 180° (Right angled at A)

∠ABC + ∠ACB = 90°

Using the same logic for △PQR, we can say that:

∠PQR + ∠PRQ + ∠QPR = 180°

∠ABC + ∠ACB + ∠QPR = 180° (∠ABC = ∠PRQ and ∠QCA = ∠CQP)

Or,

90° + ∠QPR =180° (∠ABC+ ∠ACB = 90^o)

∠QPR = 90°