Question

37. In AABC, side AB is produced to D such that
BD = BC. If B = 60° and A = 70°, prove
that (i) AD >CD and (ii) AD > AC
​

Answer 1

Answer:

ACB=50

∘

Let ∠BCD=x,∠BDC=y

x+y=60

∘

[exterior angle property]

Also, 70+50+x+y=180

∘

Given BD=BC ⇒x=y=30

∘

∠ACD=50

∘

+30

∘

=80

∘

Now, in Δ

le

ACD,CD is side opposite to ∠A & AD is side opposite to ∠C.

∵∠C>∠A

∴AD>CD[∵ side opposite to greater angle is longer ]