(1/3)

AC is tangent to circle with centre O.

Thus, $\angle ACO=90°$

In $∆AO\text{'}D$ and $∆AOC$, $\angle ADO\text{'}=\angle ACO=90°$

$\angle A=\angle A$ (Common)

$\therefore$ $∆AO\text{'}D~∆AOC \text{(By AA similarity)}$

$\Rightarrow \frac{AO\text{'}}{AO}=\frac{DO\text{'}}{CO}$

Now, $AO=AO\text{'}+O\text{'}X+XO=3r$

$\therefore$ $\frac{DO\text{'}}{CO}=\frac{r}{3r}=\frac{1}{3}$

(90°)

Join OC. Since, the tangents drawn to a circle from an external point are equal.

$\therefore$ AP = AC

In $∆$PAO and $∆$AOC, we have:

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC

$\Rightarrow ∆PAO\cong ∆AOC$ $\text{[SSS Congruency]}$

$\therefore \angle PAO=\angle CAO=\angle 1$

$\angle PAC=2\angle 1$                             ...(i)

Similarly $\angle CBQ=2\angle 2$               ...(2)

Again, we know that sum of internal angles on the same side of a transversal is 180°.

$\therefore$ $\angle PAC+\angle CBQ=180°$

$\Rightarrow 2\angle 1+2\angle 2=180° \text{[From (1) and (2)]}$

$\Rightarrow \angle 1+\angle 2=\frac{180°}{2}=90°$     ....(3)

Also $\angle 1+\angle 2+\angle AOB=180° \text{[Sum of angles of a triangle]}$

$\Rightarrow 90°+\angle AOB=180°$

$\Rightarrow \angle AOB=180°-90°$

$\Rightarrow \angle AOB=90°$

$\angle BQP=\angle BAQ (\angle \text{s in alternate segment are equal})$

$\Rightarrow \angle BQP=30° \text{...(i) (∵ }\angle BAQ=30°\text{ given)}$

As AB is a diameter, AQB is a Semicircle.

$\angle AQB=90° (\text{angle in semicircle }=90°)$

From figure, $\angle AQP=\angle AQB+\angle BQP$

$\Rightarrow \angle AQP=90°+30°=120°$

In $∆AQP,$ $\angle QPA+\angle BAQ+\angle AQP=180°$

$\Rightarrow \angle QPA+30°+120°=180°$

$\Rightarrow \angle QPA=180°-(30°+120°)$

$\Rightarrow \angle QPA=30° \text{...(ii)}$

From (i) and (ii) we get

$\angle BQP=\angle QPB=30°$

Therefore, QB = BP

Given A quad. ABCD circumscribes a circle with centre O.

To Prove: $\angle AOB+\angle COD=180°$

and $\angle AOD+\angle BOC=180°$

Join OP, OQ, OR and OS.

We know that the tangent drawn from an external point of a circle subtends equal angles at the centre.

$\Rightarrow \angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8,$

and $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360° [\angle S\text{ at a Point}]$

$\Rightarrow 2(\angle 2+\angle 3)+2(\angle 6+\angle 7)=360°$

$2(\angle 1+\angle 8)+2(\angle 4+\angle 5)=360°$

$\Rightarrow \angle 2+\angle 3+\angle 6+\angle 7=180°$

$\angle 1+\angle 8+\angle 4+\angle 5=180°$

$\Rightarrow \angle AOB+\angle COD=180°$

$\angle AOD+\angle BOC=180°$

Join OC. Since, the tangents drawn to a circle from an external point are equal.

$\therefore$ AP = AC

In Δ PAO and Δ AOC, we have:

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC

$\Rightarrow$ Δ PAO $\cong$ Δ AOC [SSS Congruency]

$\therefore$ $\angle$PAO = $\angle$CAO = $\angle$1

$\angle$PAC = 2 $\angle$1                              ...(1)

Similarly $\angle$CBQ = 2 $\angle$2               ...(2)

Again, we know that sum of internal angles on the same side of a transversal is ${180}^o$.

$\therefore$ $\angle$PAC + $\angle$CBQ = ${180}^o$

$\Rightarrow$2 $\angle$1 + 2 $\angle$2 = ${180}^o$ [From (1) and (2)]

$\Rightarrow$ $\angle$1 + $\angle$2 = ${180}^o/2={90}^o$                           ...(3)

Also $\angle$1 + $\angle$2 + $\angle$AOB = 180° [Sum of angles of a triangle]

$\Rightarrow$ 90° + $\angle$AOB = ${180}^o$

$\Rightarrow$ $\angle$AOB = ${180}^o-{90}^o$ $\Rightarrow$ $\angle$AOB = ${90}^o$.

We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.

Since tangents to a circle from an external point are equal in length,

$\therefore$ AP = AS, BP = BQ, CR = CQ and DR = DS

Adding, we get

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

$\Rightarrow$ AB + CD = AD + BC

But AB = CD [opposite sides of ABCD]

and BC = AD

$\therefore$ AB + CD = AD + BC $\Rightarrow$ 2 AB = 2 BC

$\Rightarrow$ AB = BC

Similarly AB = DA and DA = CD

Thus, AB = BC = CD = AD

Hence ABCD is a rhombus.