Given ABC is a triangle in which \(AD\bot BC\) .
To show AC ² = AB² + BC² - 2BC.BD.

proof In \(\triangle ADC\),
As, \(AD\bot BC\)
AC² = AD² + DC²  [by Pythagoras theorem]
AD² = AC² - DC²... (i)
In \(\triangle ABD\)
AB² = BD² + AD²  [by Pythagoras theorem]
AD² = AB² - BD²  ... (ii)
From Eq.(i) and Eq.(ii),
AC² - DC² = AB² - BD²
AC² = AB² + DC² - BD²
AC² = AB² + (BC - BD)²
- BD² [aS, DC = BC - BD]
= AB² + BC² + BD² - 2 . BC. BD - BD²
AC² = AB² + BC² - 2BC.BD.

Given \(\triangle PQR\) is a right angled triangle \(\angle Q\) = 90°, QS = SR.

To show PR²  = 4 PS²  - 3 PQ²
Proof \(\triangle PQR\),
PR²  = PQ² + OR²  [by Pythagoras theorem]  ... (i)
Now, QS = SR  [given]
QR = SR = \(\frac{1}{2}\) QR
QR = 2QS      ... (ii)
Put QR = 2QS in Eq.(i),
then, PR² = PQ² + (2QS)² = PQ² + 4QS²
In \(\triangle PQS\),
PS² = PQ² + QS²
\(\Rightarrow \)  QS² = PS² - PQ²    ... (iii)
\(\Rightarrow \)  PR² = PQ² + 4 (PS² - PQ²) [from Eq.(iii)]
= PQ² + 4 PS² - 4 PQ²
\(\therefore \)  PR²  = 4 PS²  - 3 PQ²

In right angled \(​​\triangle PSQ\) and \(\triangle PSR\) ,
PQ² = PS² + QS²... (i)
and PR² = PS² + SR²... (ii)
On adding Eqs.(i) and (ii), we get
PQ² + PR² = 2 PS² + QS² + SR²
= 2 (QS x SR) + QS² + SR²
[\(\because \)  PS² = QS x RS, given]

= (QS + SR)² = QR²
[\(\because \)  a² + b² + 2ab = (a + b)²]
\(\Rightarrow \)  PQ² + PR² = QR²
Hence, \(\triangle PQR\) is a right angled triangle right angled at P.

Given, \(DE\parallel BC\) , DE = 6 cm and BC = 12 cm
In \(\triangle ABC\) and \(\triangle ADE\),
\(\angle ABC=\angle ADE\)  [corresponding angles]
\(\angle ACB=\angle AED\) [corresponding angles]
and \(\angle A=\angle A\)  [common angle]
\(\therefore \triangle ABC\sim \triangle ADE\)  [by AAA similarity criterion]
We know that, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ADE \right) }{ ar\left( \triangle ABC \right) } =\frac { { \left( DE \right) }^{ 2 } }{ { \left( BC \right) }^{ 2 } } =\frac { { \left( 6 \right) }^{ 2 } }{ { \left( 12 \right) }^{ 2 } } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }\)
\(\Rightarrow \frac { ar\left( \triangle ADE \right) }{ ar\left( \triangle ABC \right) } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }=\frac { 1 }{ 4 } \)
Let \(ar\left( \triangle ADE \right) =k\) , then \(ar\left( \triangle ABC \right) =4k\)
Now, \(ar\left( \triangle DECB \right) =ar\left( \triangle ADC \right) - ar\left( \triangle ADE \right) \)
= 4k - k = 3k
\(\therefore \)  Required ratio = \(ar\left( \triangle ADE \right) : ar\left( \triangle DECB \right) \)
= k : 3k = 1 : 3

Given, \(\triangle ABC\sim \triangle QRP\), AB = 18 cm, BC = 15 cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle QRP \right) } =\frac { { \left( BC \right) }^{ 2 } }{ { \left( RP \right) }^{ 2 } } \)
But \(\frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle QRP \right) } =\frac { 9 }{ 4 } \)  given]
\(\Rightarrow \frac { { \left( 15 \right) }^{ 2 } }{ { \left( RP \right) }^{ 2 } } =\frac { 9 }{ 4 } \)  [\(\because \) BC = 15 cm]
\(\Rightarrow { \left( RP \right) }^{ 2 }=\frac { 225\times 4 }{ 9 } =100\)
\(\therefore \)  RP = 10 cm
[taking positive square root]

Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post. 
From the figure, you can see that DE is the shadow of the girl. Let DE be x metres.

Now, BD = 1.2 m \(\times\) 4 = 4.8 m.
Note that in \(\Delta\) ABE and \(\Delta\) CDE,
\(\angle\)B and \(\angle\)D  (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground)
and \(\angle\)E = \(\angle\)E   (Same angle)
So, \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\) (AA similarity criterion)
Therefore, \(\frac{\mathrm{BE}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{CD}}\)
i.e.,   \(\frac{4.8+x}{x}=\frac{3.6}{0.9} \quad\left(90 \mathrm{~cm}=\frac{90}{100} \mathrm{~m}=0.9 \mathrm{~m}\right)\)
i.e., 4.8 + x = 4x
i.e.,  3x = 4.8
i.e.,  x = 1.6
So the shadow of the girl after walking for 4 seconds is 1.6 m long.

Given, AB = DE, BC = EF and AC = DF
\(\therefore \triangle ABC\sim \triangle DEF\)  [by SSS similarity criterion]
So, x = 87°, y = 58° and z = 35°
[since, corresponding angles of similar triangles are equal]

Given \(\triangle ABC\) and \(\triangle DBC\) have same base BC and lie on the same side of BC.
To prove  \(QR\parallel AD\)
Construction Join QR and AD.

Proof Since, \(PQ\parallel AB\)
So, \(\frac { CP }{ PB } =\frac { CQ }{ QA } \)  [by basic proportionality theorem] ... (i)
\(\because PR\parallel BD\)
So, \(\frac { CP }{ PB } =\frac { CR }{ RD } \)   [by basic proportionality theorem]   ... (ii)
From Eqs. (i) and (ii), \(\frac { CQ }{ QA } =\frac { CR }{ RD } \)
Hence, \(QR\parallel AD\)
[by converse of basic proportionality theorem]

Given, in \(\triangle ABC\), \(DE\parallel BC\)
By Thales theorem, we get
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

\(\Rightarrow \frac { 4x-3 }{ 3x-1 } =\frac { 8x-7 }{ 5x-3 } \)
[\(\because \)  AD = 4x - 3, DB = 3x - 1, AE = 8x - 7, EC = 5x - 3]
\(\Rightarrow \) (4x - 3)(5x - 3) = (8x - 7)(3x - 1)
\(\Rightarrow \)  20x² - 12x + 9 - 15x = 24x² - 21x - 8x + 7
\(\Rightarrow \)  4x² - 2x - 2 = 0
\(\Rightarrow \)  2x² - x - 1 = 0     [dividing both sides by 2]
\(\Rightarrow \)  2x² - 2x + x - 1 = 0   [by splitting the middle term]
\(\Rightarrow \)  2x(x - 1) + 1 (x - 1) = 0
\(\Rightarrow \)  (2x + 1) (x - 1) = 0   \(\therefore x=-\frac { 1 }{ 2 } \) or x = 1
If \(x=-\frac { 1 }{ 2 } \), then AD = \(4\times -\frac { 1 }{ 2 } -3=-5<0\)  [not possible]
Hence, x = 1 is the required value.

Given \(ST\parallel QR\), PS/SQ = 3/5 and PR = 28 cm
By using basic proportionality theorem, we get
\(\frac { PS }{ SQ } =\frac { PT }{ TR } \Rightarrow \frac { PS }{ SQ } =\frac { PT }{ PR-PT } \)
\(\Rightarrow \frac { 3 }{ 5 } =\frac { PT }{ 28-PT } \)
\(\Rightarrow 3\left( 28-PT \right) =5PT\)
\(\Rightarrow 84=5PT+3PT\Rightarrow PT=\frac { 84 }{ 8 } \)
\(\Rightarrow PT=\) 10.5 cm
Hence, the length of PT is 10.5 cm.

Let us join AC to intersect EF at G

AB || DC and EF || AB (Given)
So, EF || DC (Lines parallel to the same line are parallel to each other)
Now, in \(\Delta\) ADC,
EG || DC (As EF || DC)
So, \(\frac{AE}{ED}=\frac{AG}{GC}\)
Similarly, from  \(\Delta\)CAB,
\(\begin{aligned} & \frac{C G}{A G}=\frac{C F}{B F} \\ \end{aligned}\)
\(\begin{aligned} & \frac{A G}{G C}=\frac{B F}{F C} \end{aligned}\)
Therefore, from (1) and (2),
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}\)

Given, in \(\triangle ABC\), \(DE\parallel AC\)   [given]
So,  \(\frac { BE }{ EC } =\frac { BD }{ DA } \)  ... (i)
[ by basic proportionality theorem]
Also, \(DC\parallel AP\)    [given]
So, \(\frac { BC }{ CP } =\frac { BD }{ DA } \)  ... (ii)
[ by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac { BE }{ EC } =\frac { BC }{ CP } \)

Since, the height h is measured vertically, so \(\angle EDA\) is a right angle. We assume that the net (i.e. CB) is vertical.
Here, \(\triangle ADE\) and \(\triangle ABC\) are similar as they are equiangular.
\(\therefore \frac { DE }{ BC } =\frac { AD }{ AB } \)   [ Corresponding sides of similar triangles are proportional]
\(\Rightarrow \frac { h }{ 0.9 } =\frac { 18 }{ 6 } \Rightarrow \frac { h }{ 0.9 } =3\)
h = 0.9 x 3 = 2.7 m
Hence, the height at which the ball should be hit, is 2.7 m.

Given, \(\triangle ABC\sim \triangle EDF\)
Also, AB = 5 cm, AC = 7 cm, DF = 15 cm
and DE = 12 cm                         .... (i)
Since, \(\triangle ABC\sim \triangle EDF\)
\(\therefore \frac { AB }{ ED } =\frac { AC }{ EF } =\frac { BC }{ DF } \)
[ ∵ Corresponding sides of similar triangles are proportions]
\(\Rightarrow \frac { 5 }{ 12 } =\frac { 7 }{ EF } =\frac { BC }{ 15 } \)  [from Eq. (i)]

On taking first and second terms, we get
\(\frac { 5 }{ 12 } =\frac { 7 }{ EF } \Rightarrow EF=\frac { 7\times 12 }{ 5 } \) = 16.8 cm
On taking first and third terms, we get
\(\frac { 5 }{ 12 } =\frac { BC }{ 15 } \Rightarrow BC=\frac { 5\times 15 }{ 12 } \) = 6.25 cm
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

Given, AB = 6.5 cm and PQ = 10.4 cm

Since, \(\triangle ABC\sim \triangle PQR\)
\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AC }{ PR } =\frac { 6.5 }{ 10.4 } =\frac { 65 }{ 104 } \)
[ \(\because \) Corresponding sides of similar triangles are proportional]
i.e. \(AB=\frac { 65 }{ 104 } PQ,BC=\frac { 65 }{ 104 } QR,AC=\frac { 65 }{ 104 } PR\)
Also given, perimeter of \(\triangle ABC\) = 60
\(\therefore AB+BC+AC=60\)
\(\Rightarrow \frac { 65 }{ 104 } \left( PQ+QR+PR \right) =60\)
\(PQ+QR+PR=\frac { 60\times 104 }{ 65 } \) = 96 cm
Hence, periemeter of \(\triangle PQR\) is 96 cm.