Question

Good morning ❤️..

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.


u have a gr8 challenge ...


all the best ... ♥️♥️

Answer 1

hey guys!!! ♥️♥️♥️

____________________________

Given :- ABCD is a quadrilateral circumscribing circle whose centre is O .

To prove :- (i) angle AOB + angle COD = 180°

(ii) angle BOC + angle AOD = 180°

construction = join OP, OQ , OR & OS .

Proof :-- AP = AS
BP = BQ
CQ = CR
DR = DS

In ∆ BOP & ∆ OBQ

OP = OQ ( Radii of same circle )
OB = OB (common )

BP = BQ (above)

∆OPB ≈ ∆ OBQ (by SSS )..

Angle 1 = angle 2 (by cpct)..

Similarly , angle 3 = angle 4
angle 5 = angle 6
angle 7 = angle 8

therefore, angle 1 + 2 + 3+4+5+6+7+8 = 360°

angle 1+1 +4+4+ 5+5++8+8 = 360°

2(angle 1+4+5+8) = 360°

angle 1 + 4 + 5 + 8+ = 180°

angle (1+5) + (4+8) = 180°

angle AOB + angle COD = 180°

Similarly , we can prove that

angle BOC + angle AOD = 180°...

hence proved !!!

♥️♥️♥️♥️♥️♥️♥️

Answer 2

Hope it'll help you.....