Question

Let PS is the bisector of QPR and PT perpendicular to QR. Show that TPS = (q - r)
and Draw its diagram.​

Answer 1

Given; PS is bisector of angle QPR

PT is perpendicular to QR

To Prove; angle TPS = 1/2 (angle Q - angle R)

Solutio; We know that angle QPS = 1/2 angle P

(1) angle Q + angle QPT = 90 degree

angle QPT = 90 degree -angle Q (2) On putting equations (1) and (2)

angle TPS = angle QPS - angle QPT

= 1/2 angle P - ( 90 - angle Q )

= 1/2 angle P - 90 + angle Q

= 1/2 angle P - 1/2 (angle Q + angle P + angle R) + angle Q

= 1/2 angle P - 1/2 angle Q - 1/2 angle P - 1/2 angle R + angle Q

= -1/2 angle Q - 1/2 angle R + angle Q (+ve and -ve 1/2 angle P got cancled )

by takig LCM in angle Q we get

-1/2 angle Q + 2/2 angle Q - 1/2 angle R

(-1/2 + 2/2) angle Q - 1/2 angle R

= 1/2 angle Q - 1/2 angle R

angle TPS = 1/2 ( angle Q -1/2 angle R ) (Hence proved)

i hope it will help u mate

Answer 2

Answer:

Given

△PQR , PS is bisector of ∠P and PT is perpendicular to QR .

To prove

∠TPS=12(∠Q−∠R)

Proof

In △TPS ,

∠TPS+∠PTS+∠ PST=180°

∠TPS+90°+(∠SPR+∠SRP)=180°

(PT is perpendicular to QR and angle PST is exterior angle of triangle PRS)

∠TPS+90°+12∠P+∠R=180°

(PS is bisector)

∠TPS+90°+12(180°−∠Q−∠R)+∠R=180° (Angle sum property of triangle)

∠TPS+90°+90°−12∠Q−12∠R+∠R=180°

∠TPS+180°−12∠Q+12∠R=180°

∠TPS=180°−180°+12∠Q−12∠R

∠TPS=12(∠Q−∠R)

Proved