A quadrilateral ABCD is drawn to circumscribe a circle .prove that AB+CD=AD+BC
Answers
Answer:
Step-by-step explanation:to prove:
AB+CD=AD+BC
Proof :
AP=AS
PB=BQ
CQ=CR
RD=DS (the length of tangent drawn from an external point to a circle are equal)
AP+PB+RD+CR=AS+BQ+DS+CQ
=>AB+CD=AD+BC
Hence proved
Given:
A quadrilateral ABCD is circumscribed in a circle.
To Find:
AB + CD = AD + BC
Solution:
Firstly connecting all the points of the circle to the quadrilateral
The length of the tangents drawn from an external point to a circle is equal.
DR = DS ..(i)
BP = BQ ..(ii)
AP = AS ..(iii)
CR = CQ ..(iv)
Since they are tangents on the circle from points D, B, A, and C respectively.
Using (i), (ii), (iii), (iv) add the LHS and RHS of the above equations separately.
DR + BP + AP + CR = DS + BQ + AS + CQ
Now, we rearrange the terms,
(DR + CR) + (BP + AP ) + (CQ + BQ) + (DS + AS)
Then simplifying the terms
CD + AB = BC + AD
AB+CD=AD+BC
Hence proved, if a quadrilateral ABCD is drawn to circumscribe a circle then AB+CD=AD+BC.