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Question
If the sides of a quadrilateral ABCD touch a circle, then____.
If the sides of a quadrilateral ABCD touch a circle, then____.
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Solution
The correct option is B AB+CD=BC+AD
Given - A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q, R, and S respectively.
To Prove - AB+ CD = BC+AD
Proof -
By Theorem- Tangents drawn from an external point to a circle are equal in lengths.
∵ AP and AS are the tangents to the circle from external points A.
∴ AP=AS ....(i)
Similarly, we can prove that,
BP = BQ ....(ii)
CR=CQ ....(iii)
DR = DS ....(iv)
On adding (i), (ii), (iii) and (iv), we get:
AP+BP+CR+DR=AS+BQ+CQ+DS
⇒ [(AP+BP)+(CR+DR)] = [(AS+DS)+(BQ+CQ)]
⇒ AB+CD=AD+BC
Hence, AB+CD=BC+AD
Given - A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q, R, and S respectively.
To Prove - AB+ CD = BC+AD
Proof -
By Theorem- Tangents drawn from an external point to a circle are equal in lengths.
∵ AP and AS are the tangents to the circle from external points A.
∴ AP=AS ....(i)
Similarly, we can prove that,
BP = BQ ....(ii)
CR=CQ ....(iii)
DR = DS ....(iv)
On adding (i), (ii), (iii) and (iv), we get:
AP+BP+CR+DR=AS+BQ+CQ+DS
⇒ [(AP+BP)+(CR+DR)] = [(AS+DS)+(BQ+CQ)]
⇒ AB+CD=AD+BC
Hence, AB+CD=BC+AD
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