Question

a quadrilateral ABCD is drawn to circumscribe a circle.prove that AB+CD=BC+AD​

Answer 1

Answer:

Step-by-step explanation:

proof-

let AB touches the circle at P, BC at Q, DC at R and AD at S.

then PB= PQ(length of tangents drawn from an external point are always equal).

Similarly,

QC=RC

AP=AS

DS=DP

Now,

AB+CD=AP+PB+DR+RC =AS+QB+DS+CQ=AS+DS+QB+CQ

=AD+BC

hence proved.

Answer 2

Answer:

AB + CD = BC + AD

Step-by-step explanation: Using the property of tangent.

In Quadrilateral ABCD and Circle with centre O,

Aa = Ad ⇒ eq(1)

Ba = Bb ⇒ eq(2)

Cc = Cb ⇒ eq(3)

Dc = Dd ⇒ eq(4)

(Tangents drawn from an external point to a circle are equal)

Now, Adding eq 1,2,3 & 4 we have,

Aa + Ba + Cc + Dc = Ad + Bb + Cb + Dd

(Aa + Ba) + (Cc + Dc) = (Ad + Dd) + (Cb + Bb)

AB + CD = AD + BC                                          (From Diagram = Aa + Ba = AB

                                                                            & similarly others)

I hope this explaination & diagram will help you!! : )