In a figure, PQRS is a parallelogram and angle SPQ=60 degreee. If the bisectors of angle P and angle Q meet at A on RS, prove that A is the midpoint of RS...

We have ∠ P = 60°

and ∠ P + ∠ Q = 180° ⇒∠ Q = 120°. (sum of interior angles on the same side of transversal)

AP is the angle bisector

∴ ∠SPA = ∠APQ = 30° (a)

similarly ∠AQR = ∠AQP = 60°

PQ is parallel to SR, and PA is a transversal

∴∠SAP = ∠APQ = 30° (alternate angles)

∴∠SAP = ∠SPA (from a)

Now in ΔPAS,

∠SAP = ∠SPA

∴ SA = PS (by isosceles triangle property) ... (1)

Similarly AR=RQ ... (2)

also PS = RQ (sides of parallelogram) ... (3)

by (1), (2), and (3)

SA = AR

∴ A is the mid point of SR.

**
**