OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have 
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

In \(\triangle\)PST and ​​\(\triangle\)PQR
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+3 } =\frac { 2 }{ 5 } ,\frac { PT }{ PR } =\frac { 2 }{ 2+3 } =\frac { 2 }{ 5 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is is common
Therefore, by SAS similarity,
\(\triangle\)PST~\(\triangle\)PQR

In \(\triangle\)ACB, 
PQ || CB
Using Basic Proportionality theorem, we have
\(\frac{A Q}{A B}=\frac{A P}{A C}\)
Again in \(\triangle\)ACD PR || CD
Using Basic Proportionality theorem
\(\frac{A P}{A C}=\frac{A R}{A D}\)
From (1) and (2)
\(\frac{A Q}{A B}=\frac{A P}{A C}=\frac{A R}{A D} \)
Thus we have \(\frac{A R}{A D}=\frac{A Q}{A B} \)

We have AB = 56.cm, AD = 14. cm, AC = 72. cm and AE = 18.cm.
BD = AB - AD = 5.6 –1.4 = 4.2 cm
and EC = AC – AE = 7.2–1.8 = 5.4 cm
\(\frac { AD }{ DB } =\frac { 1.4 }{ 4.2 } =\frac { 1 }{ 3 } \) and \(\frac { AE }{ EC } =\frac { 1.8 }{ 5.4 } =\frac { 1 }{ 3 } \)
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Therefore, by converse of Basic Proportionality Theorem, we have DE is parallel to BC. Hence proved.

(i) In ABC and ADE <A is common
\( \frac{A E}{E C}=\frac{2}{3 \frac{1}{2}}=\frac{\frac{2}{7}}{2}=\frac{2 \times 2}{7}=\frac{4}{7} \)
\(\frac{A D}{D B}=\frac{3}{5} \)
\(Here\ \frac{4}{7} \pm \frac{3}{5} \)
\(\frac{A E}{E C} \neq \frac{A D}{D B} \)
The corresponding sides are not proportional.
ABC and ADE are not similar
(ii) In CPQ and CAB <C is common
<PQC = 180^o - 110^o = 70^o
[ <PQC and <PQB are liner pair of angles]
<ABC = 70^o
<BAC = <QPC
[ sum of three angles of a triangle are 180]
<PCQ = 180^o - (< QPC + 70^o)
<ABC = <PQC = 70^o
<C common and <BAC = <QPC
By AAA similarity criteria, <ABC <PQC
Corresponding sides are Proportional
\(\frac{A B}{P Q} =\frac{B C}{Q C} \)
\(\frac{5}{\sqrt{x}} =\frac{6}{3} \)
[BC = BQ + QC = 3 + 3 = 6]
\(x=\frac{5}{6} \times 3=2.5\)
x = 2.5

In Is \(\triangle\)ABC ~ \(\triangle\)PQR
\(\frac { PQ }{ AB } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } ;\frac { QR }{ BC } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Since \(\frac { 1 }{ 2 } \neq \frac { 2 }{ 5 } ,\frac { PQ }{ AB } \neq \frac { QR }{ BC } \)
The corresponding sides are not proportional.
Therefore \(\triangle\)ABC  is not similar to \(\triangle\)PQR.

We are given a right triangle ABC right angled at B.
We need to prove that AC² = AB² + BC²
Let us draw \(BD\bot AC\)
Now,\(\Delta ADB\sim \Delta ABC\)
\(\cfrac { AD }{ DB } =\cfrac { BC }{ AC } \)
(sides are proportional)
Also,
\(\Delta BDC\sim \Delta ABC\)
\(\cfrac { CD }{ BC } =\cfrac { BC }{ AC } \)
CD·AC = BC²     ..(2)
Adding (1) and (2)
AD .AC + CD . AC = AB²+ BC²
AC(AD + CD) = AB² + BC²
AC.AC = AB² + BC²
AC = AB² + BC²

From \(\Delta ADC\) we have
AC² = AD² + CD²       ....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB² = AD² + BD²     ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² - AC² = BD² - CD²
AB² + CD² = BD² + AC²

Let the first aeroplane starts from O and goes upto A towards north, (Distance=Speed × time)
where \(OA=\left( 100\times \frac { 3 }{ 2 } \right) km=1500km\)
Let the second aeroplane starts from O at the same time and goes upto B towards west,
where \(OB=(1200\times \frac { 3 }{ 2 } )=1800km\)
The required distance to be found is BA.
In right angled triangle AOB, AOB, AB² = OA² + OB²
AB² = (1500)² + (1800)² = 100² (15² +18²)
= 100² x 549 = 100² x 9 x 61
\(AB=100\times 3\times \sqrt { 61 } =300\sqrt { 61 } kms.\)

In \(\triangle\)BPA, we have DE||AP By Basic Proportionality Theorem,
We have \(\frac { BC }{ CP } =\frac { BD }{ DA } \)   ..(i)
In \(\triangle\)BCA, we have DE||AC By Basic Proportionality Theorem,
we have,
\(\frac { BE }{ EC } =\frac { BD }{ DA } \)         ..(2)
From (1) and (2) we get, \(\frac { BE }{ EC } =\frac { BC }{ CP } \), Hence proved.

Given, Speed = 1.2 m/s,
time = 4 seconds
Distance = speed x time
= 1.2 x 4
= 4.8 m
Let x be the length of the shadow after 4 seconds
\(\Delta ABE\sim \Delta CDE,\frac { BE }{ DE } =\frac { AB }{ CD } \) gives \(\frac { 4.8+x }{ x } =\frac { 3.6 }{ 0.9 } =\frac { 3.6 }{ 0.9 } =4\) (since 90 cm = 0.9 m)
4.8 + x = 4x, gives 3x = 4.8 so, x = 1.6m
The length of his DE = 1.6m