9. In parallelogram ABCD, the bisector of angle
A meets DC at P and AB = 2AD.
Prove that :
(i) BP bisects angle B. (i) Angle APB = 90°.
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Answers
Given :
- In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2AD.
To prove :
- (i) BP bisects angle B.
- (ii) Angle APB = 90°.
Solution :
i]
(1) In || gm ABCD AP is bisector of ∠A
∠1 = ∠2
(2) AB || DC and AP is the transversal,
∠2 = ∠3 (Alternate angles)
(3) ∠1 = ∠3 From (1) and (2)
(4) In ΔADP, we have
∠1= 2∠3 Proved in (3)
DP=AD
[Sides opposite to equal angles are equal]
AD = DP
(5) AB = 2AD........ (Given)
AB/2 = AD
AD = AB/2
(6) AB || DC
[Opposite sides of || gm are equal]
(7) AD = DC/2
(8) DP = DC/2 From (4) and (7)
P is the mid-point of DC.
DP = PC
(9) AD = BC
[Opposite sides of || gm are equal]
(10) DP = BC From (4) and (9)
(11) PC = BC From (8) and (10)
(12) In ΔBCP, we have
PC = BC Proved in (11)
∠6 = ∠5
[Angles opposite to equal sides are equal.]
(13) AB || DC
[ABCD is a || gm and BP is the transversal ]
∠4 = ∠5 (alternate ∠s)
(14) ∠4 = ∠6 From (12) and (13)
BP is bisector of ∠B.
ii]
(15) AD || BC
[Opposite sides of gm ABCD and AB
is the transversal]
∠A + ∠B = 180°
[Sum of co-interior angles = 180°]
=> ½(∠A+ ∠B) = ½ x 180
=> ∠APB = 90°