Question

ΔPQR  is an isosceles triangle with PQ = PR. QR is extended to S and ST is drawn perpendicular to QP produced, and SN is perpendicular to PR produced. Prove that QS bisects ∠TSN.

Answer 1

In ΔPQR, let ∠PQR = x
PQ = PR
⇒ ∠PQR = ∠PRQ = x   ........(i)
In ΔRNS,
∠NRS = ∠PRQ = x    .........(vertically opposite angles)
∠RNS = 90°     ...(given)
∠NSR + ∠RNS + ∠NRS = 180°
∠NSR + 90° + x  = 180
∠NSR = 90° - x  .........(ii)
Now in Quadrilateral PTRS
∠PTS = 90°    ...(given)
∠TPR = ∠PQR + ∠PRQ = 2x  ....(exterior angle to triangle PQR)
∠PRS = 180°  -  ∠PRQ = 180° - x    ...(QRS is a st. Line)
∠PTS + ∠TRP + ∠PRS + ∠TSR = 360°   ...(angles of a quad. = 360°)
90° + 2x + 180° - x + ∠TSR = 360°
∠TSR = 90° - x  ..........(iii)
From (ii) and (iii)
∠TSR = ∠NSR
Therefore, QS bisects ∠TSN.