Q86 of 93 Page 6

In fig, ∠ Q >∠ R and M is a point on QR such that PM is the bisector of ∠ QPR. If the perpendicular from P on QR meets QR at N, then prove that
∠ MPN = (∠ Q - ∠ R).
          

Given: Δ PQR, PM is the bisector of ∠ QPR such that ∠ QPM  = ∠ MPR and PN  ^  QR.
To prove: ∠ NPM = (∠ Q -∠ R)
Proof: In Δ NPM, ∠ PNM = 90°
∴ ∠ NPM = 90° - ∠ PMN [Sum of the angles of a triangle is 180° ]……….(i)
In Δ PMR, ∠ PMN = ∠ MPR + ∠ PRM (Ex. Angle of a triangle is sum of the interior opposite angles]
But ∠ MPR = ∠ QPM = [MP is the bisector of ∠ QPR]
∴ ∠ PMN = + ∠ PRM ………………….(ii)
In Δ PQR, ∠ P + ∠ Q + ∠ R = 180° {Sum of the three angles of a triangle is 180° ]
+ + = = 90°
                       ∴ = 90° - ( +)………………..(iii)
Substituting (iii) in (ii),
∠ PMN = 90° - ( + ) + ∠ R
         = 90° - - + ∠ R
         = 90° - + ……………………..(iv)
Substituting (iv) in (i),
∠ NPM = 90° - [90° -   + ]
         = 90° - 90° + -
         = -
         = (∠ Q -∠ R)
∴ ∠ NPM = (∠ Q -∠ R)

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(ii) If the arms of an angle are respectively perpendicular to the arms of another angle, then the angles will be supplementary: 87 In figure m and n are two plane mirrors perpendicular to each other. Prove that the incident ray CA is parallel to reflected ray BD.
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