In ΔMAO and ΔCAO,
AM = AC
(Tangents drawn from external point to a
circle are equal) AO = AO    (Common)
and    OM = OC (radii of circle)
Therefore, by SSS congruent condition
ΔAMO ≅ ΔAOC
⇒ ∠OAM = ∠OAC
⇒     ∠MAB = 2 ∠OAC
or    ∠MAB = 2 ∠OAB
Similarly, we can prove
∠NBA = 2 ∠OBC
or    ∠NBA = 2 ∠OBA
Now, ∠MAB + ∠NBA = 180°
(Consecutive interior angles of the same side of transversal)
⇒ 2∠OAB + 2∠OBA = 180°
⇒ (∠OAB + ∠OBA) = 90°
Now, in Δ AOB
(∠OAB + ∠OBA + ∠AOB = 180°
(angle sum property of triangle)
Δ    90° + ∠AOB = 180°
⇒    ∠AOB = 90.

Since, the tangent at any point of a circle is perpendicular to radius through the point of contact.
Therefore, ∠OPQ = 90°
It is given that OQ = 5 cm
and    PQ = 4 cm

In right ΔOPQ, we have
OQ² = OP² + PQ²
[Using Pythagoras Theorem]
OP² = OQ² – PQ²
⇒ OP² = (5)² – (4)²
= 25 – 16 = 9
⇒ OP = 3 cm
Hence, the radius of the circle is 3 cm.

Let O be the common centre of the two concentric circle.
Let PQ be a chord of the larger circle which touches the smaller circle at M.
Join OM and OP.
Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,
∠OMP = 90°
Now,
In ΔOMP, we have
OP² = OM² + PM²
[Using Pythagoras theorem]
⇒ (5)² = (3)² + PM²
⇒ 25 = 9 + PM²
⇒ PM² = 16
⇒ PM = 4 cm
Since, the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
PM = MQ = 4 cm
∴ PQ = 2 PM = 2 x 4 = 8 cm
Hence, the required length = 8 cm.

Let AMB and CND be two parallel tangents to a circle with centre O. Join OM and ON. Draw OP || AB.
Now, AM || PO
⇒ ∠AMO + ∠POM = 180°
[Consecutive interior angles]
⇒    90 + ∠POM = 180°
⇒    ∠POM = 90°
Similarly, ∠PON = 90°
∴ ∠POM + ∠PON = 90° + 90° =180°
Hence, MON is a straight line passing through O.

Since, tangents drawn from an exterior point to a circle are equal in length.

∴ AP = AS    ...(i)
BP = BQ    ...(ii)
CR = CQ    ...(iii)
and DR = DS    ...(iv)
Adding (i), (ii), (iii) and (iv), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR)
= (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence, AB + CD = BC + DA.