Question

in Δpqr pq=pr,ps ia the angle bisector of angleP then show that Ps perpendicularQr

Answer 1

In ΔPQR,

PQ = PS

∴ΔPQR is an isosceles Δle

⇒ ∠Q = ∠R [Angles opp. to the equal sides are equal]

PS bisector of ∠P

∴ ∠2 = ∠1 [Let ∠P be equal to ∠2 + ∠1]

In ΔPQR,

    ∠P + ∠Q + ∠R  = 180°            [ASP]

     ∠ P + ∠R + ∠R = 180°     [As ∠Q = ∠R]

                        ∠P = 180° - 2∠R          

                        ∠P = 180° - 2∠R

                          ----    ----------------

                            2            2

∴∠1 = 90° - ∠R  →1

In ΔPSR,

∠1 + ∠PSR + ∠R = 180°                         [ASP]

90° - ∠R  + ∠PSR + ∠R = 180°       [From 1]

                 90° + ∠PSR = 180°

∴∠PSR = 90°

∴PS is ⊥r to QR.

∴Hence proved.