Question

In the given figure, PS perpendicular QR, SP perpendicular MN, and angle MPQ = angle NPR. Prove that triangle PQS congruent to triangle PRS​

Answer 1

Step-by-step explanation:

Given:

In ΔPQR, PA is the bisector of ∠QPR and PM ⊥ QR. To prove: ∠APM = 1 2 12 (∠Q – ∠R)

proof:

Suppose ∠APR = ∠1, ∠APM = ∠2 and ∠QPM = ∠3 ∠1 = ∠2 + ∠3 …

(i) [As PA is the bisector of ∠QPR] In ΔPMR, ∠MPR + ∠PMR + ∠PRM = 180°

(angle sum property of a triangle)

⇒ (∠1 + ∠2) + 90° + ∠PRM = 180°

⇒ (∠1 + ∠2) + ∠PRM = 90° …

(ii) Similarly in ΔPQM, ∠3 + ∠Q = 90° …

(iii) [As ∠PMQ = 90°]

From (ii) and (iii), we get ∠1 + ∠2 + ∠PRM = ∠3 + ∠Q

⇒ ∠1 + ∠2 + ∠R = ∠3 + ∠Q

⇒ ∠1 + ∠2 – ∠3 = ∠Q – ∠R

⇒ (∠1 – ∠3) + ∠2 = ∠Q – ∠R (as ∠Q > ∠R)

Using relation (i), we get ∠2 + ∠2 = ∠Q – ∠R

⇒ 2∠2 = ∠Q – ∠R

⇒ ∠2 = 1 2 12 (∠Q – ∠R)

⇒ ∠APM = 1 2 12 (∠Q –

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