Question

3A. Complete any one out of two activities:
03
1.
Given:
DABCD is cyclic. ZDCE is an
D
A А
exterior angle of DABCD. To prove
ZDCE = BAD complete the following
C с
E
(1)
proof by filling the boxes
ZDCE + <BCD - LUI
Linear pair of
DABCD is cyclic
BAD+O- 180° (Theorem of IN
(2)
from (1) and (2) we get
ZDCE +
+ BOD
ZDCE =​

Answer 1

Answer:

In given figure ABCD is a cyclic quadrilateral.

We know that, sum of the opposite angles in a cyclic quadrilateral is 180

∘

So, ∠BCD + ∠BAD = 180

∘

⇒ 125

∘

+ ∠BAD = 180

∘

⇒ ∠BAD = 180

∘

- 125

∘

∴ ∠BAD = 55

∘

---- (1)

Since, ∠ABD is an angle in semi-circle.

∴ ∠ABD = 90

∘

--- (2)

In △ABD,

∠ADB + ∠ABD + ∠BAD = 180

∘

∠ADB + 90

∘

+ 55

∘

= 180

∘

(From 1 and 2)

∴ ∠ADB = 35

∘