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Question

Prove that opposite sides of a quadrilatral circumscribing a ciecle subtend supplementary angle at the centre of the circle.
1462271_f0386aabcb7f41cbad21c47ba939edc5.png


Solution

Given: Let ABCD be the quadrilateral circumscribing the circle with centre O.
ABCD touches the circle at points P, Q, R and S.

To prove: Opposite sides subtends supplementary angles at centre.
i.e., $$\angle AOB+\angle COD=180^o$$

& $$\angle AOD+\angle BOC=180^o$$

Construction: Join OP, OQ, OR and OS.


Proof: Let us rename the angles

In $$\Delta AOP$$ and $$\Delta AOS$$,
AP$$=$$AS
(Length of tangents drawn from external point to a circle are equal)

AO$$=$$AO (common)

OP$$=$$OS (both radius)

$$\Delta AOP\cong \Delta AOS$$(SSS congruence rule)

$$\angle AOP=\angle AOS$$(CPCT)

i.e., $$\angle 1=\angle 8$$ …………….$$(1)$$

Similarly, we can prove
$$\angle 2=\angle 3$$ ……………….$$(2)$$
$$\angle 5=\angle 4$$ ………………..$$(3)$$
$$\angle 6=\angle 7$$ ……………….$$(4)$$

Now,
$$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^o$$
(sum of angles round a point is $$360^o$$)

$$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 1=360^o$$

$$\angle 1+\angle 2+\angle 5+\angle 6=\dfrac{360^o}{2}$$

$$(\angle 1+\angle 2)+(\angle 5+\angle 6)=180^o$$

$$\angle AOB+\angle COD=180^o$$

Hence  both angles are supplementary.

similarly, we can prove
$$\angle BOC+\angle AOD=180^o$$

Hence proved.

1225135_1462271_ans_3b9674ab52384176a7f32185c291160a.png

Mathematics

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