127
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Show SolutionHide Solution↓
Let ABCD be a quadrilateral circumscribing a circle with centre O. Let the sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.
Join OA, OB, OC, OD, OP, OQ, OR, OS.
In $\triangle AOP$ and $\triangle AOS$:
OP = OS (Radii of the same circle)
OA = OA (Common side)
AP = AS (Tangents from an external point A)
So, $\triangle AOP \cong \triangle AOS$ (SSS congruence criterion)
$\Rightarrow \angle 1 = \angle 2$ (CPCTC)
Similarly, we can prove:
$\triangle BOQ \cong \triangle BOP \Rightarrow \angle 3 = \angle 4$
$\triangle COR \cong \triangle COQ \Rightarrow \angle 5 = \angle 6$
$\triangle DOR \cong \triangle DOS \Rightarrow \angle 7 = \angle 8$
The sum of all angles around the centre O is $360^\circ$.
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ$ (since $\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8$)
$\angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$
Now, consider angles subtended by opposite sides at the centre:
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8)$
Since $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$
$\angle AOB + \angle COD = (\angle 1 + \angle 3) + (\angle 5 + \angle 7)$ (This step is incorrect in the provided solution, it should be $\angle 1+\angle 4$ and $\angle 5+\angle 8$)
Let's re-evaluate: $\angle AOB = \angle 1 + \angle 4$, $\angle COD = \angle 5 + \angle 8$
$\angle BOC = \angle 3 + \angle 6$, $\angle DOA = \angle 2 + \angle 7$
We know $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$.
So, $2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ \Rightarrow \angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$.
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8) = (\angle 1 + \angle 3) + (\angle 5 + \angle 7) = 180^\circ$.
Similarly, $\angle BOC + \angle DOA = (\angle 3 + \angle 6) + (\angle 2 + \angle 7) = (\angle 3 + \angle 5) + (\angle 1 + \angle 7) = 180^\circ$.
Thus, opposite sides subtend supplementary angles at the centre._a_1.png)
Join OA, OB, OC, OD, OP, OQ, OR, OS.
In $\triangle AOP$ and $\triangle AOS$:
OP = OS (Radii of the same circle)
OA = OA (Common side)
AP = AS (Tangents from an external point A)
So, $\triangle AOP \cong \triangle AOS$ (SSS congruence criterion)
$\Rightarrow \angle 1 = \angle 2$ (CPCTC)
Similarly, we can prove:
$\triangle BOQ \cong \triangle BOP \Rightarrow \angle 3 = \angle 4$
$\triangle COR \cong \triangle COQ \Rightarrow \angle 5 = \angle 6$
$\triangle DOR \cong \triangle DOS \Rightarrow \angle 7 = \angle 8$
The sum of all angles around the centre O is $360^\circ$.
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ$ (since $\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8$)
$\angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$
Now, consider angles subtended by opposite sides at the centre:
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8)$
Since $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$
$\angle AOB + \angle COD = (\angle 1 + \angle 3) + (\angle 5 + \angle 7)$ (This step is incorrect in the provided solution, it should be $\angle 1+\angle 4$ and $\angle 5+\angle 8$)
Let's re-evaluate: $\angle AOB = \angle 1 + \angle 4$, $\angle COD = \angle 5 + \angle 8$
$\angle BOC = \angle 3 + \angle 6$, $\angle DOA = \angle 2 + \angle 7$
We know $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$.
So, $2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ \Rightarrow \angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$.
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8) = (\angle 1 + \angle 3) + (\angle 5 + \angle 7) = 180^\circ$.
Similarly, $\angle BOC + \angle DOA = (\angle 3 + \angle 6) + (\angle 2 + \angle 7) = (\angle 3 + \angle 5) + (\angle 1 + \angle 7) = 180^\circ$.
Thus, opposite sides subtend supplementary angles at the centre.