We know that, tangents drawn from an exterior point to a circle are equal in length.
AD = AF = x cm
BD = BE = y cm
CE = CF = z cm
Given, AB = 8 cm
⇒ AD + BD = 8cm
⇒ x + y = 8
BC = 10 cm ⇒ BE + CE = 10 cm
⇒ y + z = 10
and CA = 12 cm ⇒ CF + AF = 12 cm
⇒ z + x = 12
On adding Eqs. (i), (ii) and (iii), we get
2(x + y + z) =30
⇒ x + y + z = 15
On subtracting Eq. (ii) from Eq. (iv), we get
x =15 - 10 = 5
On subtracting Eq. (iii) from Eq. (iv), we get
y = 15-12 = 3
On subtracting Eq. (i) from Eq. (iv), we get
z=15-8=7
AD = xcm = 5 cm
BE= ycm = 3 cm
and CF = z cm = 7 cm
Hence, the length of AD, BE and CE are 5 cm, 3 cm and 7 cm, respectively.

5 cm

r = 14 cm

We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P (see Fig). We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore, by Theorem.
OP \(\perp\) AB
Now AB is a chord of the circle C₁ and OP \(\perp\) AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
i.e., AP = BP

Secant

Let OP intersect AB at a point C. Here, PA and PB are the two tangents from a point P lying outside the circle, to the circle with centre O. 

\(\angle \)APO = \(\angle \)BPO [∵ O lies on the bisector of ZAPB] 
Now, in \(\triangle\)ACP and \(\triangle\)BCP, we have 
AP = BP [tangents from an external pointl 
PC = PC [commonl]
\(\angle \)APO = \(\angle \)BPO [proved above] 
⇒ \(\triangle\)ACP ≅ \(\triangle\)BCP   [by SAS congruence axiom]
⇒ AC = BC   [c.p.c.t.] 
and  \(\angle \)ACP = \(\angle \)BCP 
= \(1\over2\) x 180° = 90° [c.p.c.t.]
Hence, OP is the perpendicular bisector of AB. 

We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P see fig. We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then \(\angle\)OQP and \(\angle\)ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore,   \(\Delta\)OQP \(\cong\)\(\Delta\) ORP   (RHS)
This gives  PQ = PR    (CPCT)

64 cm

\({ 100 }^{ \circ }\)

Given: A circle with centre O. AB is the diameter of this circle. I is tangent to the circle. AH and BK are perpendicular to I from A and B at H  and K respectively. 
To prove: AH + BK = AB 
Proof: AH and HP are tangents from the external point H 
ஃ  AH = HP        .....(i)
and BK, KP are tangent from the external point K
ஃ   BK = KP      ......(ii)
Adding (i) and (ii) we get
AH + BK = HP + PK = HK ......(iii)
AB ⊥  AH
AB ⊥ BK
[Tangent makes 90° angle with radius at the point of contact]
⇒ \(\angle \)1 = \(\angle \)2 = 90°
Given that AH ⊥ i ⇒ \(\angle \)3 = 90°
and BK ⊥ i ⇒ \(\angle \)4 = 90°
∵ \(\angle \)1 = \(\angle \)2 = \(\angle \)3 = \(\angle \)4 = 90°​​​​​​​
⇒ AHKB is a rectangle
⇒  AB = HK        ......(iv)
[Opposite sides of a rectangle are equal]
From (iii) and (iv) ⇒ PH + PK = AB
AH + BK = AB from (i) and (ii) Hence proved.

r₁ = 5 cm, r₂ = ?,
AB = 24 cm
∵ AB is tangent to circle
C(C, r₁) at C
ஃ OC ⊥ AB
In circle C(O, r₂), AB is a chord and OC ⊥ AB
ஃ AC = BC
[∵ Perpendicular from the centre bisects the chord]
In right \(\triangle\)OCA
OC² + AC² = AC²
⇒ 5² + (12)² = (r₂)²
⇒ 25 + 144 = (r₂)² ⇒ (r₂)² = 169
r₂ = 13 cm

AB = 24 cm
AC = 10 cm
In righta angled ABC
BC² = AB² + AC²
= 24² + 10²
= 276
⇒ BC =- 26 cm
Let r be the radius of the incircle
⇒ OP ⊥ AB, OQ ⊥ AC and OR ⊥ BC
[Tangent 90° angle with the radius at the point of contact]
OP = OQ = OR
[Incentre of a triangle is equidisation from its sides]
ar (\(\triangle\)ABC) = ar (\(\triangle\)AOB) + ar (\(\triangle\)BOC) + ar(\(\triangle\)AOC)
\(1\over2\)AB x AC = \(1\over2\)AB x OP + \(1\over2\)AC x OQ + \(1\over2\) x BC x OR
\(1\over2\) x 24 x 10 = \(1\over2\)[24 x r x +10 x r + 26 x r]
​​⇒ 120 = r[24 + 10 + 26]
⇒ 120 = 30r ⇒ r = 4 cm

In right CAB
BC² = AC² + AB² = 8² + 6²
BC?² = 100 ⇒ BC = 10 cm

Area of \(\triangle\)CAB = \(1\over2\) x AB x AC
= \(1\over2\) x 6 x 8 = 24 cm²
Area of \(\triangle\)AOB  = \(1\over2\) x AB x OP
= \(1\over2\)x 6 x r = 3r cm²
Area of \(\triangle\)AOC = \(1\over2\) x AC x OT
= \(1\over2\) x 8 x r = 4r cm²
Area of \(\triangle\)BOC = \(1\over2\)x BC x OS
= \(1\over2\) x 10 x 10 x r = 5r cm²
Now, area \(\triangle\)AOB + area \(\triangle\)AOC + area \(\triangle\)BOC = Area \(\triangle\)ABC
⇒ 3r + 4r + 5r = 24 ⇒ 12r = 24
ஃ r = 2 cm

Given.  AB and CD are two tangents to a circle and AB || CD.
Tangent BD intercepts an angle BOD at the centre.
To prove. \(\angle \)BOD = 90°
Construction. Join OQ, OB and OR.
Proof.   OP ⊥ BD.
[A tangent at any point of a circle is perpendicular to the radius through the point of contact.]
In right angled \(\triangle\) s OQB and OPB
\(\angle \)1 = \(\angle \)2,
Similarly in right angled \(\triangle\)s OPD and ORD
\(\angle \)3 = \(\angle \)4
ஃ   \(\angle \)BOD = \(\angle \)1 + \(\angle \)3 = \(1\over2\)[2\(\angle \)1 + 2\(\angle \)3] =\(1\over2\)(\(\angle \)1 + \(\angle \)1 + \(\angle \)3 + \(\angle \)3)
= \(1\over2\)(\(\angle \)1 + \(\angle \)2 + \(\angle \)3 + \(\angle \)4 ) = \(1\over2\)(180°) = 90°