Question

in the given figure a quadrilateral ABCD is drawn to circumscribe a circle with the centre O in such a way that the sides ab BC CD and D A touch the circle at the points P Q R and S respectively prove that AB + CD is equal to BC + CA​

Answer 1

Step-by-step explanation:

The figure shows that the tangents drawn from the exterior point to a circle are equal in length.

As DR and DS are tangents from exterior point D so, DR = DS---- (1)

As AP and AS are tangents from exterior point A so, AP = AS---- (2)

As BP and BQ are tangents from exterior point B so, BP = BQ---- (3)

As CR and CQ are tangents from exterior point C so, CR = CQ---- (4)

Adding the equation 1,2,3 & 4, we get

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

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

CD+AB=DA+BC

AB+CD=BC+DA

Hence proved.

Answer 2

Answer:

This answer for you are first question

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