# In the given figure PQRS is aparalelogram and line segements PA and RB bisects the angles p and r respectivley show that PA PARALLEL TO RB

2
by tanveeram23shuja

2016-01-09T21:02:04+05:30
Let ∠P = 2x degrees
.`. ∠R = 2x degrees (as opp angles of a parallelogram are equal)

.`. ∠APB = ∠ ARB = 2x/2 = x degrees (as PA and RB bisects ∠P and ∠ R                                                                              respectively [given])

A line segment PQ is drawn through the points P and Q such that it bisects the angles ∠APB and ∠ARB.

Let PQ be a transversal.

Alternate angles ∠PRB and ∠APR are formed.

∠PRB = ∠ APR = x /2 degrees (as PQ bisects angles ∠APB and ∠ARB)

.`. Alternate angles are equal.

.`. PA || RB [Proved]
2016-01-12T07:43:42+05:30
Let ∠P = 2x degrees .`. ∠R = 2x degrees (as opp angles of a parallelogram are equal) .`. ∠APB = ∠ ARB = 2x/2 = x degrees (as PA and RB bisects ∠P and ∠ R                                                                              respectively [given]) A line segment PQ is drawn through the points P and Q such that it bisects the angles ∠APB and ∠ARB. Let PQ be a transversal. Alternate angles ∠PRB and ∠APR are formed. ∠PRB = ∠ APR = x /2 degrees (as PQ bisects angles ∠APB and ∠ARB) .`. Alternate angles are equal. .`. PA || RB [Proved]