Question

In the given figure, ABCD is a parallelogram
Pis a point in CD such that AD = DP = PC.
Prove that:
(1) AP bisects A
(i) BP bisects B
(ii) DAP + CBP = APB
​

Answer 1

Answer:

Step-by-step explanation:

ABCD is parallelogram AD=DP=PC=CB

To check

(i) AP bisect angle A

(ii) BP bisects angle B

(iii) ∠DAP−∠CBP=∠APS

⇒ (i) ABCD is parallelogram

AB∣∣DC⇒AB∣∣DP and AB∣∣PC.

Now, InΔADP

AD=DP

∴∠DPA=∠DAP

∠3=∠1 __(1) [Angles opposite to equal sides of a triangle are equal]

also, ∠3=∠2 __(2) [AP acts as transversal for AB∣∣DP]

From (1) and (2) [So, alternate interior angles are equal]

∴AP bisects angle ∠DAB

(ii) In ΔPCB,

CB=PC

∠4=∠5 __(3) [Angles opposite to equal sides of a triangle are equal]

Also, PC∣∣AB

∴PB acts as transversal.

Alternate interior angles are equal.

∠4=∠6 ___(4)

From (3) and (4),

∠5=∠6

∴PB bisects angle B.

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