Question

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.​

Answer 1

Answer:

Let ABCD be a quadrilateral circumscribing a circle with centre O.

Now join AO, BO, CO, DO.

From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]

Let ∠DAO=∠BAO=1

Also ∠ABO=∠CBO [Since, BA and BC are tangents]

Let ∠ABO=∠CBO=2

Similarly we take the same way for vertices C and D

Sum of the angles at the centre is 360

o

Recall that sum of the angles in quadrilateral, ABCD = 360

o

=2(1+2+3+4)=360

o

=1+2+3+4=180

o

In ΔAOB,∠BOA=180−(1+2)

In ΔCOD,∠COD=180−(3+4)

∠BOA+∠COD=360−(1+2+3+4)

=360

o

–180

o

=180

o

Since AB and CD subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Figure in attachment pls see

Hope it helps you

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Answer 2

Question :-

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle

Answer :-

Let ABCD be a quadrilateral circumscribing a circle with centre O. A circle touches the sides of a quadrilateral at points E, F ,G and H.

$\star\;{\underline{\boxed{\pmb{\green{\frak{\;To \:proove\; : -}}}}}}$ $\angle{AOB}$ + $\angle{COD}$ = 180°

and $\angle{AOD}$ + $\angle{COD}$ = 180°.

$\star\;{\underline{\boxed{\pmb{\green{\frak{Construction}}}}}}$

• join OH , OF ,OE and OG.

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ Proof Using the property , two tangents from the external point to a circle subtend equal angles at the centre.

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\angle{1}$ = $\angle{2}$

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\angle{3}$ = $\angle{4}$...(1)

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\angle{5}$= $\angle{6}$

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\angle{7}$= $\angle{8}$

We know the sum of all angles subtenided at point O is 360°

$\therefore$ $\angle{1}$ +$\angle{2}$ +$\angle{3}$ +$\angle{4}$ +$\angle{5}$+ $\angle{6}$ + $\angle{7}$+ $\angle{8}$ = 360°

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ 2( $\angle{2}$ +$\angle{3}$ +$\angle{6}$ +$\angle{7}$ = 360°

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\angle{1}$+ $\angle{4}$ +$\angle{7}$+ $\angle{8}$ = 360° ...[From Equation ..(1)

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ ($\angle{2}$+ $\angle{4}$) + ( $\angle{6}$+ $\angle{7}$) = 180°

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ ($\angle{1}$+ $\angle{8}$)+ ($\angle{4}$+ $\angle{5}$) = 180°

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\angle{AOB}$+ $\angle{COD}$ = 180°

$\:\:\:\:\:\:\:\:\:\huge\rightarrow$ $\angle{AOD}$+ $\angle{BOC}$ = 180°

$\Huge\sf\implies$ Hence proved