Question

3/4 ribbon is equally shared by Chinnu and mine
how much does each got?
3/4x1/2=3/8metre
can you ask questions related to everyday life like this?​

Answer 1

Given : In ||gm ABCD,

∠A = 60°

Bisector of ∠A and ∠B meet at P.

To prove :

(i) AD = DP

(ii) PC = BC

(iii) DC = 2AD

Construction : Join PD and PC

Proof : In ||gm ABCD,

∠A = 60°

But ∠A + ∠B = 180° (Sum of excutive angles)

⇒ 60° + ∠B = 180°

∴ ∠B = 1809 – 60° = 120°

∵ DC || AB

∠PAB = ∠DPA (alternate angles)

⇒ ∠PAD = ∠DPA (∵ ∠PAB = ∠PAD)

∴ AB = DP

(PA is its angle bisector, sides opposite to equal angles)

(ii) Similarly, we can prove that ∠PBC = ∠PCB (∵ ∠PAB = ∠BCA alternate angles)

∴ PC = BC

(iii) DC = DP + PC

= AD + BC [From (i) and (ii)]

= AD + AB = 2AB (∵BC = AD opposite sides of the ||gm)

Hence DC = 2AD