show that if a circle touches all the four sides of a quadrilateral ABCD at points P,Q,R,S then AB+CD=BC+DA.
Answers
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:
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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
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