Question

Prove that the parallelogram circumscribing a circle is a rhombus.
Or, ∆ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 8 cm and 6 cm respectively. Find AB and AC.
Figure

Answer 1

Given: A parallelogram ABCD circumscribes a circle with centre O.
Thus, AB = CD and AD = BC
To Prove : AB = BC = CD = AD
Proof :
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
∴ AP = AS .......(i) [tangents from A]
BP = BQ .......(ii) [tangents from B]
CR = CQ .......(iii) [tangents from C]
DR = DS .......(iv) [tangents from D]

∴ AB + CD = AP + BP + CR + DR
= AS + BQ + CQ + DS [From (i), (ii), (iii) and (iv)]
= (AS + DS) + (BQ + CQ)
= AD + BC
Thus, AB + CD = AD + BC
⇒ AB + AB = AD+ AD
⇒ 2AB = 2AD
⇒ AB = AD
∴ CD = AB = AD = BC
Hence, ABCD is a rhombus.

OR
Let the given circle touch the sides AB and AC of the triangle at point F and E, respectively, and let the length of the line segment AF be x.

In ΔABC, BF = BD = 6 cm (Tangents on the circle from point C)
CE = CD = 8 cm (Tangents on the circle from point B)
AE = AF = x cm (Tangents on the circle from point A)
AB = AF + FB = (x + 6) cm
BC = BD + DC = (8 + 6) = 14 cm
CA = CE + EA = (8 + x ) cm
∴ Perimeter of triangle (2s) = AB + BC + CA = (x + 8 + 14 + 6 + x) = (28 + 2x) cm
Semi-perimeter (s) = (14 + x) cm
Now,
$\mathrm{Area}\mathrm{of}△\mathrm{ABC}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
$=\sqrt{\left\{14+x\right\}\left\{\left(14+x\right)-14\right\}\left\{\left(14+x\right)-\left(6+x\right)\right\}\left(14+x\right)-\left(8+x\right)}$
$=\sqrt{\left(14+x\right)\left(x\right)\left(8\right)\left(6\right)}=4\sqrt{3\left(14x+x^2\right)}$
Area of ΔOBC = $\frac{1}{2}\times \mathrm{OD}\times \mathrm{BC}=\frac{1}{2}\times 4\times 14=28$
Area of ΔOCA = $\frac{1}{2}\times OE\times \mathrm{AC}=\frac{1}{2}\times 4\times \left(8+\mathrm{x}\right)=16+2\mathrm{x}$
Area of ΔOAB = $\frac{1}{2}\times OF\times \mathrm{AB}=\frac{1}{2}\times 4\times \left(6+\mathrm{x}\right)=12+2\mathrm{x}$
Again,
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
$4\sqrt{3\left(14x+x^2\right)}=28+16+2x+12+2x$
$\Rightarrow 4\sqrt{3\left(14x+x^2\right)}=56+4x$
$\Rightarrow 4\sqrt{3\left(14x+x^2\right)}=4(14+x)$
$\Rightarrow \sqrt{3\left(14x+x^2\right)}=14+x$
Squaring both the sides, we get:
3(14x + x²) = (14 + x)²
⇒ 42x + 3x² = 196 + 28x + x²
⇒ 2x² +14x -196 = 0
⇒ x² + 7x - 98 = 0
⇒ x² + 14x - 7x - 98 = 0
⇒ x (x + 14) - 7( x + 14) = 0
⇒ (x - 7)(x + 14) = 0
⇒ (x +14 ) = 0 or (x − 7) = 0
Therefore, x = − 14 or x = 7
However, x = − 14 is not possible as the length of the sides cannot be negative.
Therefore, x = 7
Hence, AB = x + 6 = 7 + 6 = 13 cm
AC = 8 + x = 8 + 7 = 15 cm