10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that \[AB+CD=BC+DA\].

Answer:

We have, AB, BC, CD and DA are the tangents touching the circle at P, Q, R and S respectively

Now, \[AP=AS,BP=BQ,CR=CQ\] and \[DR=DS\].

On adding we get

\[AP+BP+CR+DR=AS+BQ+CQ+DS\]

\[AB+CD=AD+BC\] Hence Proved.

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10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

question_answer In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that \[AB+CD=BC+DA\].

In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that \[AB+CD=BC+DA\].