Question

A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

Answer 1

Answer:

The figure given is:

From this figure we can conclude a few points which are:

(i) DR = DS

(ii) BP = BQ

(iii) AP = AS

(iv) CR = CQ

Since they are tangents on the circle from points D, B, A, and C respectively.

Now, adding the LHS and RHS of the above equations we get,

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

By rearranging them we get,

(DR+CR) + (BP+AP) = (CQ+BQ) + (DS+AS)

By simplifying,

AD+BC= CD+AB

Answer 2

Given:-

Let ABCD be the quadrilateral circumscribing the circle with centre O. The quadrilateral touches the circle at point P,Q,R and S.

To prove:-

AB+CD+AD+BC

Proof:-

As we know that, length of tangents drawn from the external point are equal.

Therefore,

AP=AS-------(1)

BP=BQ------(2)

CR=CQ------(3)

DR=DS------(4)

Adding equation (1),(2),(3) and (4), we get

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

(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)

⇒AB+CD=AD+BC

Hence proved.