In Fig. 14.34, ABCD is a parallelogram in which ∠A=60°. If the bisectors of ∠A and ∠B meet at P, prove that AD=DP, PC=BC and DC=2AD.
Answers
Proved below.
Step-by-step explanation:
Given:
Here ABCD is a parallelogram.
Also the bisectors of ∠A and ∠B meet at P and ∠A=60°.
To prove:
AD=DP, PC=BC and DC=2AD.
Proof:
∠A + ∠B = 180° [ ABCD is a parallelogram]
60 + ∠B = 180° [given]
∠B = 180° - 60°
∠B = 120 [1]
We also know that opposite angles of a parallelogram are equal.
∴ ∠D = ∠B
∴ ∠D = 120° [from Eq (1)]
We are also given that AP as the bisector of ∠A.
∴ ∠DAP = ∠BAP [1]
Similarly,
∠ABP = ∠PBA [2]
We have DC║AB
∠DPA =∠PAB
∠DAP = ∠DPA [from Eq (1)]
Thus, sides opposite to equal angles are equal.
AD = DP [3]
Similarly, DC║AB
∠CPB = ∠PBA
∠CPB = ∠PBC [from Eq (2)]
Thus, sides opposite to equal angles are equal.
PC = BC [4]
Since ABCD is a parallelogram, we have AD = BC and AB = CD [5]
Also, DC = DP + PC
DC = AD + BC [from Eq (3) and (4)]
DC = AD + AD [from Eq (5)]
DC = 2AD
Hence proved.