Question

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

Answer 1

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{answer}}}}}}$

DR=DS ।$\pink{Tangents\:from\:same\:point..}$।

AP=AS ।$\pink{Tangents\:from\:same\:point..}$।

BP=BQ ।$\pink{Tangents\:from\:same\:point..}$।

RC=CQ ।$\pink{Tangents\:from\:same\:point..}$।

Adding All these-

=>DR+AP+BP+RC = DS+AS+BQ+CQ

$\blue{DR+RC}$+$\green{AP+BP}$=$\purple{DS+AS}$+$\orange{BQ+CQ}$

=>$\huge\blue{CD}$+$\huge\green{AB}$=$\huge\purple{AD}$+$\huge\orange{BC}$

$hence\:proved$

Step-by-step explanation:

$<marquee>$Hope it helps...

Answer 2

Answer:

since ABCD is a parallelogram

AB=CD.......1

BC=AD......2

DR=DS (tangent on the circle from point D)

BP=BQ (tangent circle to point B)

CR=CQ (point C)

AP=AS (point A)

adding all this equation

DR + CR + BP + AP + DS + CQ + BQ + AS

[DR+CR] + [BP+AP] = [DS+AS] + [CQ+BQ]

CD+AB = AD+BC

AB=BC......3

comparing 1 , 2 and 3

AB = BC = CD = DA

Hence ABCD is a rohmbus