Statement
If a straight line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Proof

Given
\(\text { In } \triangle A B C, \frac{A D}{D B}=\frac{A E}{E C}\)
To prove : DE \(\|\) BC
Construction : If DE is not parallel to BC, draw DF \(\|\) BC .

Statement :
The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle
Proof

Given : In ΔABC,AD is the internal bisector
To prove : \(\frac{A B}{A C}=\frac{B D}{C D}\)
Construction : Draw a line through C parallel to AB. Extend AD to meet line through C at E

Statement
If a line touches a circle and from the point of contact a chord is drawn, the angles between the tangent and the chord are respectively equal to the angles in the corresponding alternate segments.
Proof
Given : A circle with centre at O, tangent AB touches the circle at P and PQ is a chord. S and T are two points on the circle in the opposite sides of chord PQ.

To prove : (i) \(\angle\)QPB = \(\angle\)PSQ and (ii)\(\angle\)QPA = \(\angle\)PTQ
Construction: Draw the diameter POR. Draw QR, QS and PS.

Statement
In a right angle triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Proof
Given : In \(\triangle\)ABC , \(\angle\)A = 90°
To prove : AB² +AC² = BC²
Construction : Draw AD \(\perp\) BC

Adding (1) and (2) we get
AB² +AC² = BC x BD +BC x DC
= BC(BD +DC) = BC x BC
AB² +AC² = BC² .
Hence the theorem is proved.

Statement
If a straight line through one vertex of a triangle divides the opposite side internally in the ratio of the other two sides, then the line bisects the angle internally at the vertex.
Proof

Given : ABC is a triangle. AD divides BC in the ratio of the sides containing the angles \(\angle\)A to meet BC at D.
That is \(\frac{A B}{A C}=\frac{B D}{D C}\)
To prove : AD bisects \(\angle\)A i.e. \(\angle\)1 = \(\angle\)2
Construction : Draw CE \(\|\)DA . Extend BA to meet at E.

Statement
A straight line drawn parallel to a side of triangle intersecting the other two sides, divides the sides in the same ratio.
Proof

In \(\Delta ABC\) ,D is a point on AB and E is a point on AC
To prove : \(\cfrac { AD }{ DB } =\cfrac { AE }{ EC } \) 
Construction: Draw a line DE || BC

\(\frac{A D}{D B}=\frac{A E}{E C}\)
Given AD = 8x - 7
DB = 5x - 3
AE = 4x - 3
EC = 3x - 1
\(\frac{8 x-7}{5 x-3}=\frac{4 x-3}{3 x-1}\)
(8x - 7) (3x - 1) = (4x - 3) (5x- 3)
24x² - 21x - 8x + 7 -  20x² - 15x - 12 x + 9
24x² - 20x² - 29x + 27x + 7 - 9 = 0
4x² - 2x - 2 = 0
by 2 2x² - x - 1 = 0
2x (x - 1) + 1(x - 1) = 0
(x - 1)(2x + 1) = 0
x - 1 = 0 or 2x + 1 = 0
x = 1 or \(x=-\frac{1}{2}\)
x cannot be negative
x = 1

Let \(\triangle\)ABC be the given triangle and D, E, F are points on the respectives sides AB, BC AC (or their extension).
Given AD : DB = 5 : 3;
BE : EC = 3 : 2.
Let AD = 5k₁; DB = 3k₁; BE = 3k₂; EC = 2k₂.
Now by Menelaus Theorem, we have
\( \frac{B E}{E C} \times \frac{C F}{F A} \times \frac{A D}{D B}=-1 \)
\(\frac{3 k_{2}}{2 k_{2}} \times \frac{C F}{F C-A C} \times \frac{5 k_{1}}{3 k_{1}}=-1 \)
\(\frac{3}{2} \times \frac{C F}{F C-21} \times \frac{5}{3}=-1 \)
\(\frac{C F}{F C-21}=\frac{-1 \times 2}{5} \)
\(\frac{C F}{(-C F)-21}=\frac{-2}{5}\)
5 CF = -2 [(-CF) - 21]
5CF = 2CF + 42 [FC = -CF]
5CF - 2CF = 42
3CF = 42
\(C F=\frac{42}{3}=14 \text { units. }\) 

Let MP is the tangent at p Join AB.
\(\angle M P A=\angle P B A\)          ...(1)
[Angles in the alternate segment]
\(\angle P B A=\angle A C D\)           ...(2)
[Exterior angle of a cyclic quadrilateral
ABCD is equal to its interior opposite angle]
From (1) and (2), we have
\( \angle M P A =\angle A C D \)
\(\Rightarrow \ \angle M P C =\angle P C D \)
But these are a pair of alternate interior angles.
\(M P \| C D\)

Let AB be the emu = 8 ft
CD be the pillar = 30 ft
OB be the shadow, BD be the distance between the pillar and the emu.
Draw AE || OD at a distance of 8 ft from OD.
Now, \( \angle C A E=\angle A O B \)   [corresponding angles]
\(\angle O B A=\angle A E C=90^{\circ}\)
By AA similarity criteria
\( \triangle C E A \sim \triangle A B O \)
\(\frac{C E}{A B} =\frac{E A}{B O} \)
\(\frac{22}{8} =\frac{B D}{O B} \quad[\because \mathrm{EA}=\mathrm{BD}] \)
\(\frac{11}{4} =\frac{\text { distance }}{\text { shadow }} \)
\(\text { Shadow } =\frac{4}{11} \times \text { distance }\) 

Let CP be the man whose eye level is 2 m above the ground
AT be the height of the tree
In triangles \(\triangle C B P \text { and } \triangle A B T\)
\(\angle C=\angle A=90^{\circ}\)
Since \(\mathrm{CA} \perp \mathrm{PC} \text { and } \mathrm{CA} \perp \mathrm{TA}\)
\(\angle C B P=\angle A B T \text { as } \angle C B P\) is the angle of reflection and \(\angle\)ABT is the angle of incident.
By AA similarity criteria
\( \triangle C B P \sim \triangle A B T \)
\(\therefore \ \frac{C B}{A B} =\frac{B P}{B T}=\frac{C P}{A T} \)
\(\frac{C B}{A B} =\frac{C P}{A T} \)
\(\frac{4}{20} =\frac{2}{A T} \)
\(\mathrm{AT} =\frac{2 \times 20}{4}=10 \mathrm{~m}\)
Height of the tree is 10 m.

Distance travelled by the first train in 2 hours
= 2 x 20 = 40km
Distance travelled by the second train in 2 hours
= 2 x 30 = 60km
Let the distances are represents by OB and OA respectively
Now applying Pythagoras theorem,
Distance between the trains after 2 hours is AB.
we have AB² = OA² + OB² = 60² + 40²
= 3600 + 1600
= 5200
\( A B =\sqrt{5200} \)
\(=\sqrt{2^{2} \times 2^{2} \times 5 \times 5 \times 13} \)
\(=2^{2} \times 5 \sqrt{13} \)
\(=20 \sqrt{13} \mathrm{~km}\)
Distance between the trains after 2 hrs
\(=20 \sqrt{13} \mathrm{~km}\)

To prove the points B, C, E and D lie on a same circle, it is sufficient to show that \(\angle ABC+\angle CED={ 180 }^{ 0 }\) and \(\angle ACB+\angle BDE={ 180 }^{ 0 }\) to prove opposite angles of the quadrilateral BCED are supplementary
  
In \(\Delta ABC\) we have
AB = AC AD = AE.
\(\Rightarrow AB-AD=AC-AE\)
\(\Rightarrow DB=EC\)
Thus we have
AD = AE and DB = EC. 
\(\Rightarrow \frac { AD }{ DB } =\frac { AE }{ EC }\)
\( \Rightarrow DE||AE\) [By the converse of Thales's theorem]
\(\Rightarrow \angle ABC=\angle ADE\) (corresponding angles)
\(\Rightarrow \angle ABC+\angle BDE=\angle ADE+\angle BDE\) (Adding  \(\angle BDE\) on both sides)
\(\Rightarrow \angle ABC+\angle BDE={ 180 }^{ 0 }\)
\(\Rightarrow \angle ACB+\angle BDE={ 180 }^{ 0 }\)           ...(1)
\(\left( \because AB=AC\therefore \angle ABC=\angle ACB \right) \)
Again DE IIBC
\(\Rightarrow \angle ACB=\angle AED\)
\(\Rightarrow \angle ACB+\angle CED=\angle AED+\angle CED\) (Adding \(\angle CED\) on both sides).
\(\Rightarrow \angle ACB+\angle CED={ 180 }^{ 0 }\)       \(\left( \because \angle ABC=\angle ACB \right) \)
\(\Rightarrow \angle ABC+\angle CED={ 180 }^{ 0 }\)             ...(2)
From (1) and (2),
Thus BDEC is a quadrilateral such that
\(\Rightarrow \angle ACB+\angle BDE={ 180 }^{ 0 }\) and
\(\Rightarrow \angle ABC+\angle CED={ 180 }^{ 0 }\)
\(\therefore\) BDEC is a cyclic quadrilateral.
Points B, C, E and D lie on a same circle.

In \(\Delta AOB\) OD is the bisector of \(\angle AOB\)
\(\therefore \frac { OA }{ OB } =\frac { AD }{ DB } \)          ...(1)
In \(\Delta BOC\) OE is the bisector of \(\angle BOC\)
\(\therefore \frac { OB }{ OC } =\frac { BE }{ EC } \)         ...(2)
In \(\Delta COA\) OF is the bisector of \(\angle COA\)
\(\therefore \frac { OC }{ OA } =\frac { CF }{ FA } \)        ...(3)
Multiplying the corresponding sides of (1), (2) and (3), we get
\(\frac{O A}{O B} \times \frac{O B}{O C} \times \frac{O C}{O A} \mid=\frac{A D}{D B} \times \frac{B E}{E C} \times \frac{C F}{F A}\)
\(1=\frac { AD }{ DB } \times \frac { BE }{ EC } \times \frac { CF }{ FA } \)
\(\Rightarrow \) DB x EC x FA = AD x BE x CF
 AD x BE x CF = DB x EC x FA
Hence proved.

Given AB || CD || EF
AB = 6 cm, BD = 5 cm, EF = 4 cm, CD = x cm, DE = y cm
\(\text { In } \triangle E C D \text { and } \triangle E A B\)
\( \angle C E D=\angle A E B \)            [common]
\(\angle E C D=\angle E A B \)             [corresponding angles]
\(\triangle E C D \sim \triangle E A B\)
[By AA similarity criteria]      ...(1)
\(\therefore \ \frac{E C}{E A}=\frac{C D}{A B}\)
[ Corresponding parts of similar triangles are proportional]
\(\frac{E C}{E A}=\frac{x}{6}\)            ....(2)
In \(\triangle A C D\ and\ \triangle A E F \)
\(\angle C A D=\angle E A F\)          [Common]
\( \angle A C D =\angle A E F \)        [Corresponding angles]
\(\triangle A C D \sim \triangle A E F \)      [By AA similarity]
\(\frac{A C}{A E} =\frac{C D}{E F}\)
[Corresponding parts of similar triangle are proportional]
\(\therefore \quad \frac{A C}{A E}=\frac{x}{4}\)          ....(3)
Adding (2) and (3)
\( \frac{E C}{E A}+\frac{A C}{A E} =\frac{x}{6}+\frac{x}{4} \)
\(\frac{E C+A C}{A E} =\frac{4 x+6 x}{24} \)
\(\frac{A E}{A E} =\frac{10 x}{24} \)
\(1 =\frac{10 x}{24} \)
\(\mathrm{x} =\frac{24}{10}=\frac{12}{5} \mathrm{~cm}\)
From (1) \(\triangle E C D \sim \triangle E A B\)
\( \frac{D C}{A B} =\frac{E D}{E B} \)
\(\frac{x}{6} =\frac{y}{5+y} \)
\(\because \mathrm{x}=\frac{12}{5} \Rightarrow \frac{12 / 5}{6} =\frac{y}{5+y} \)
\(\frac{12}{5 \times 6} =\frac{y}{5+y} \)
12(5 + y) = 30 y
60 + 12y = 30 y
60 = 30y - 12y
18y = 60
\( y=\frac{60}{18} \)
\(y=\frac{10}{3} \mathrm{~cm}\)

Clearly In \(\triangle\)ABC, D, E, F are points on lines BC, CA, AB respectively using Ceva's theorem, we have
\(\frac{A E}{E C} \times \frac{C D}{D B} \times \frac{B F}{F A}=1\)       ...(1)
From the diagram it is clear that
AE = 3, EC = 4, CD = 10, DB = 3, FA = 5
Substituting these values in (1)
\( \frac{3}{4} \times \frac{10}{3} \times \frac{B F}{5} =1 \)
\(B F =\frac{1 \times 4 \times 3 \times 5}{3 \times 10}=2 \mathrm{~cm}\)

Consider DABC, Then D, E, F are respective points on the sides CA, AB and Be. By construction D, E, F are collinear
By Menelaus' Theorem, \(\frac { AE }{ EB } \times \frac { BF }{ FC } \times \frac { CD }{ DA } =1\)
FC = FB + BC = BF + 6
By Pythagoras Theorem AC² = AB² + BC²
=64+36=100
\(\therefore \) AC = 10, CD = 8

\(4BF=BF+6\Rightarrow BF=2cm\)

Let \(\triangle\)ABC be B triangle points D, E, F are angular bisectors of \(\angle A, \angle B \text { and } \angle C\) respectively. By angular bisector theorem we have
\( \frac{B D}{D C}=\frac{A B}{A C} \Rightarrow \mathrm{AB}=\frac{B D \times A C}{D C} \)
\(\frac{A C}{B C}=\frac{A F}{F B} \Rightarrow \mathrm{AC}=\frac{A F \times B C}{F B} \)
\(\frac{A E}{E C}=\frac{A B}{B C} \Rightarrow \mathrm{AB}=\frac{A E \times B C}{E C}\)
From (1) and (3), we have
\(\frac{B D \times A C}{D C}=\frac{A E \times B C}{E C}\)
Now substituting (2) in (4) we have
\( \frac{B D \times\left(\frac{A F \times B C}{F B}\right)}{D C} =\frac{A E \times B C}{E C} \)
\(\frac{B D \times A F \times B C}{D C \times F B} =\frac{A E \times B C}{E C} \)
\(B D \times A F \times E C =\frac{A E \times B C \times D C \times F B}{B C} \)
\(B D \times A F \times C E =E A \times F B \times D C \)
\(\therefore \frac{B D \times A F \times C E}{E A \times F B \times D C}=1\)
Hence by Ceva's theorem we conclude that the angle bisectors of a triangle are concurrent.

Since OP is the radius and PT tangent
\(\angle O P T=90^{\circ}\)
Applying Pythagoras theorem in \(\triangle\)OPT, we have
OT² = OP² + PT²
13² = 5² + PT²
PT² = 169 - 25
PT² = 144
PT = 12cm
Since the lengths of tangents drawn from an exterior point to a circle are equal.
AP = AE = x(say)
AT = PT - AP = (12-x)cm
Since AB is the tangent to the circle at E
\( \therefore \mathrm{OE} \perp \mathrm{AB} \)
\(\ \Rightarrow \angle O E A =90^{\circ} \)
\(\\ \Rightarrow \angle A E T =90^{\circ}\)
\( \\ \mathrm{AT}^{2} =\mathrm{AE}^{2}+\mathrm{ET}^{2}\)
[Applying Pythagoras theorem in \(\triangle\)AET ]
(12 - x)² = x² +(13 -5)²
144 - 24 + x² = x² +64
24x = 144 - 64
24x = 80
3x = 10
\( x=\frac{10}{3} \mathrm{~cm} \)
\(Similarly\ B E=\frac{10}{3} \mathrm{~cm}\)
AB = AE + BE
\( =\left(\frac{10}{3}+\frac{10}{3}\right) \mathrm{cm} \)
\(A B =\frac{20}{3} \mathrm{~cm}\)

By Menelaus' Theorem, the trees P, Q, R will be collinear (lie on same straight line)
If \(\frac { BP }{ PC } \times \frac { CQ }{ QA } \times \frac { RA }{ RB } \)
Given BP = 2 m, CQ = 3 m, RA = 10 m, PC = 6 m, QA = 5 m and RB = 2 m
Substituting these values in (1) we get,
\(\frac { BP }{ PC } \times \frac { CQ }{ QA } \times \frac { RA }{ RB } =\frac { 2 }{ 6 } \times \frac { 3 }{ 5 } \times \frac { 10 }{ 2 } =\frac { 60 }{ 60 } \)
Hence the trees P, Q, R lie on a same straight line.

Given that AB = 13, AC = 14 and BC = 15
Let BD = x and DC = y
Using Ceva’s theorem, we have, \(\frac { BD }{ DC } \times \frac { CE }{ EA } \times \frac { AF }{ FB } =1\)
Substitute the values of  \(\frac { AF }{ FB } \ and \ \frac { CE }{ EA } \) in (1)
we have \(\frac { BD }{ DC } \times \frac { 5 }{ 8 } \times \frac { 2 }{ 5 } =1\)
\(\frac { x }{ y } \times \frac { 10 }{ 40 } =1\) we get \(\frac { x }{ y } \times \frac { 1 }{ 4 } \), Hence x = 4y  ...(2)
BC = BD + DC = 15 so, x + y = 15   ..(3)
From (2), using x = 4y in (3) we get, 4y + y = 15 gives 5y = 15 then y = 3
Substitute y = 3 in (3) we get, x = 12. Hence BD = 12, DC = 3.

Consider\(\triangle\)ABC. Then D, E and F are respective points on the sides CA, AB and BC. By construction D, E, F are collinear
By Menelaus’ theorem \(\frac { AE }{ EB } \times \frac { BF }{ FC } \times \frac { CD }{ DA } =1\)
By assumption, AE = EB = 2, DA = 1 and 
FC = FB + BC = BF + 3
By Pythagoras theorem,AC²=AB²+BC²=16+9=25. Therefore AC=5
and So, CD = AC – AD = 5 – 1 = 4.
Substituting the values of FC, AE, EB, DA, CD in (1)
we get, \(\frac { 2 }{ 2 } \times \frac { BF }{ BF+3 } \times \frac { 4 }{ 1 } =1\)
4BF=BF+3
4F-BF=8 therefore BF=1

Medians are line segments joining each vertex to the midpoint of the corresponding opposite sides.
Thus medians are the cevians where D, E, F are midpoints of BC, CA and AB respectively
Since D is a midpoint of BC, BD = DC so \(\frac { BD }{ DC } =1\)        ..(1)
Since, E is a midpoint of CA, CE = EA so \(\frac { CE }{ EA } =1\)       ..(2)
Since, F is a midpoint of AB, AF FB so \(\frac { AF }{ FB } =1\)       ...(3)
Thus, multiplying (1), (2) and (3) we get,
\(\frac { BD }{ DC } \times \frac { CE }{ EA } \times \frac { AF }{ FB } =1\times 1\times 1=1\)
And so, Ceva’s Theorem is satisfied.
Hence the Medians are concurrent.

Let TR = y. Since, OT is perpendicular bisector of PQ
PR = QR = 4 cm
In\(\triangle\)ORP, OP² = OR² + PR²
OR² = OP² - PR²
OR² = 5² - 4² = 25 - 16 = 9 \(\Rightarrow\) OR = 3cm
OT = OR + RT = 3 + y          ..(1)
In \(\triangle\)PRT, TP² + TR² + PR²      ..(2)
and \(\triangle\)OPT we have, OT² = TP² + OP²
OT² = (TR² + PR²) + OP² (substitute for TP² from (2))
(3 + y)² = y² + 4² + 5² (substitute for OT from (1))
9 + 6y² + 16 + 25 therefore \(y=TR=\frac { 16 }{ 3 } \)
6y = 41 - 9 we get  \(y=\frac { 16 }{ 3 } \)
From (2), TP² = TR² + PR²
\(TP2=\left( \frac { 16 }{ 3 } \right) +4^{ 2 }=\frac { 256 }{ 9 } +16=\frac { 400 }{ 9 } \)so, \(TP=\frac { 20 }{ 3 } \)

Since D and E are the points of trisection of BC,
therefore BD = DE = CE
Let BD = DE = CE = x
Then BE = 2x and BC = 3x
In right triangles ABD, ABE and ABC, (using Pythagoras theorem)
We have AD² = AB² + BD²
\(\Rightarrow \) AD² = AB² + x²        ... (1)
\(\text { In rt } \triangle \mathrm{ABE} \text {, }\) AE² = AB² + BE²
= AB² + (2x)²
\(\Rightarrow \) AE² = AB² + 4x²   ..(2)
\(\text { In rt } \triangle \mathrm{ABC}, \quad A C^{2}=A B^{2}+B C^{2}\)
AC² = AB² + (3x)²
AC² = AB² + 9x²           ..(3)
8AE² - 3AC² - 5AD² = 8(AB² + 4x²) - 3 
(AB² + 9x²) - 5(AB² + x²)
8AB² + 32x² - 3AB² - 27x²- 5AB² - 5x²
8AE² - 3AC² - 5AD² = 0
8AE² = 3AC² + 5AD²
Hence proved.

We have
DB = 3CD (given)
\(\because \) BC = BD + DC
\(\Rightarrow\) BC = 3CD + CD
\(\Rightarrow\) BC = 4CD
\(\Rightarrow CD=\frac { 1 }{ 4 } BC\)

Since \(\triangle\)ABD is a right triangle right angled at D.
\(\therefore { AB }^{ 2 }=AD^{ 2 }+{ BD }^{ 2 }\)
Similarly \(\Delta ACD\) is a right triangle right angled at D.
AC² = AD² + CD⁸
Subtracting equation (3) from equation (2)
We get AB² - AC² = BD² - CD²
\(\Rightarrow { AB }^{ 2 }-{ AX }^{ 2 }=\left( \frac { k }{ 4 } BC \right) ^{ 2 }-\left( \frac { 1 }{ 4 } BC \right) ^{ 2 }\)
\((from\ (1)CD=\frac { 1 }{ 4 } BC,BD=\frac { 3 }{ 4 } BD)\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\frac { 9 }{ 16 } { BC }^{ 2 }-\frac { 1 }{ 16 } { BC }^{ 2 }\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\frac { 1 }{ 2 } { BC }^{ 2 }\)
\(\Rightarrow 2{ AB }^{ 2 }=2{ AC }^{ 2 }+{ BC }^{ 2 }\)
Hence it is proved.

Clearly the ladder AC make a right triangle with the wall AB force at a distance BC. \(\angle\)B - 90^o

By Pythagoras theorem
AC² = AB² + BC²
5² = 4² + BC²
BC² = 25 - 16
BC² = 9
BC = 3m
If C moves 1.6 m towards the wall BC becomes
3m - 1.6m = 1.4m
Now in \(\triangle\)ABC
AC² = AB² + BC²
5² = AB2 +(1.4)²
25 - 1.96 = AB²
AB² = 23.04
AB = 4.8m
The new height of the wall = 4.8 m
Difference =4.8 - 4 = -0.8m
The ladder would be placed 0.8 m upward the wall.

Let A be the starting position and 'B' be the final position. C be the position south of A at 34 m distance.
Clearly C = 90^o in \(\triangle\)ACB
AC² + CB² = AB²
[By Pythagoras theorem]
34² + 41² = AB²
1156 + 1681 = AB²
2837 = AB²
AB = 53.26 m
Distance from A to B through the pond = 53.26 m
Distance through C = 34m + 41 m = 75 m
Difference = 75 - 53.26 = 21.74 m
21.74 m would be saved if it is possible to walk through the pond.

Let Sarah's house is at A and James's house is at 'B' from the picture.
Distance between Sarah's house to James house through Street B and C
= 1.5 miles + 2 miles = 3.5 miles
Distance through street C is AC² = AB² + BC²

AC² = = (1.5)² + (2),
= 2.25 + 4
= 6.25
\(A C=\sqrt{6.25}=2.5\)
AC = 2.5 miles
Difference between two paths = 3.5 - 2.5 = 1 mile
Direct path along C street is 1 mile shorter

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
\(AB=100\times 3\times \sqrt { 61 } =300\sqrt { 61 } kms.\)

Since, \(\triangle\)QAQC is a right triangle at C, AQ² = AC² + QC²  ...(1)
Also, \(\triangle\)BPC is a right triangle at C, BP² = BC²+ CP²        ...(2)
\(\triangle\) ABCC is a right triangle at C, AB² = AC² + BC²              ....(3)
From (1) and (2), AQ²+  BP² = AC²+ QC² + BC² + CP²
4(AQ² + BP²) = 4AC² + 4QC² + 4BC² + 4CP²
= 4AC² + (2QC)²+ ABC² + (2CP)²
=  4AC² + BC² + 4BC² + AC² (Since P and Q are mid points)
= 5(AC² + BC²) (From equation (3))
4(AQ² + BP²) = 5AB²

Given: A quadrilateral ABCD in which AB = AD and the bisectors of \(\angle\)BAC and \(\angle\)CAD meet the sides BC and CD at E and F respectively.
To prove: EF || BD
Construction: Join AC, BD and EF.
Proof: In \(\triangle\)CAB, AE is the bisector of \(\angle\)BAC
\(\therefore \frac{A C}{A B}=\frac{C E}{B E}\)            ......(1)
In \(\triangle\)ACD , AF is the bisector of \(\angle\)CAD
\( \frac{A C}{A D} =\frac{C F}{D F} \)
\(\Rightarrow \ \frac{A C}{A B} =\frac{C F}{D F}\)     ......(2)
From (1) and (2) we get
\( \frac{C E}{B E}=\frac{C F}{D F} \)
\(\frac{C E}{E B}=\frac{C F}{F D}\)
Thus in \(\triangle\)CBD, E and F divide the sides CB and CD respectively in the same ratio.
By the converse of Thales theorem, we have EF || BD.

Given that PS is the bisector of \(\angle\)P of \(\triangle\)PQR
\(\frac{P Q}{P R}=\frac{Q S}{S R}\) [By Angle Bisector Theorem]
Adding 1 on both the sides
\( \frac{P Q}{P R}+1 =\frac{Q S}{S R}+1 \)
\(\frac{P Q+P R}{P R}=\frac{Q S+S R}{S R} \)
\(\frac{P Q+P R}{P R} =\frac{Q R}{S R} \)           ..(2)
In \(\triangle\)RST and \(\triangle\)RQP we have
\( \angle S R T=\angle Q R P=\angle R \)
\(\angle Q P R=\angle S T R=90^{\circ}\)
By AA criterion for similarity, we have
\(\triangle R S T \sim \triangle R Q P \)
\(\frac{R S}{R Q}=\frac{S T}{Q P} \)
\(\frac{Q P}{S T}=\frac{Q R}{R S}\)             ...(2)
From (1) and (2)
\(\frac{Q P}{S T}=\frac{P Q+P R}{P R}\)
PQ \(\times\) PR = ST(PQ + PR)

In ABC , we have DE || BC
\(\frac{A B}{A D}=\frac{A C}{A E}\) [By Thales Theorem]        ..(1)
In ADC, we have
\(\frac{A D}{A F}=\frac{A C}{A E}\) [By Thales Theorem]        .....(2)
From (1) and (2) we get
\(\frac{A B}{A D}=\frac{A D}{A F}\)
AD² = AB x AF

Given: ABCD is a trapezium in which DC || AB and EF || AB
To prove that \(\frac { AE }{ ED } =\frac { BF }{ FC } \)
Construction : join AC meeting EF at G
Proof:
In ADC, we have
EG || DC
\(\Rightarrow \frac{A E}{E D}=\frac{A G}{G C}\) [By Thales theorem]               ...(1)
In ABC , we have
\(\frac{A G}{G C}=\frac{B F}{F C}\) [By Thales theorem]        ....(2)
From (1) and (2), we get
\(\frac{A E}{E D}=\frac{B F}{F C}\)

Given AB = 12 cm, BC = 6 cm
In \(\triangle\)APQ and \(\triangle\)QRC
(PB || QR, corresponding angles)
( PQ || BR, corresponding angles)
By AA criterion of similarity, we have
\(\triangle A P Q \sim \triangle Q R C \)
\( \Rightarrow \frac{A P}{Q R}=\frac{P Q}{R C}=\frac{A Q}{Q C} \)
\(\frac{A P}{Q R}=\frac{P Q}{R C} \)
\(\frac{P Q}{A P}=\frac{R C}{Q R} \)
Now in \(\triangle\)APQ and \(\triangle\)ABC , we have
By AA criterion of similarity,
we have \(\triangle\)APQ - \(\triangle\)ABC
\(\frac{A P}{A B}=\frac{P Q}{B C}=\frac{A Q}{A C} \)
\(\frac{A P}{A B}=\frac{P Q}{B C} \)
\(\frac{P Q}{A P}=\frac{B C}{A B} \)
\(\frac{P Q}{A P}=\frac{6}{12} \)
Since PQRB is a rhombus, PQ = QR = RB = PB
\(\frac{P Q}{A B-P B}=\frac{6}{12} \)
\(\frac{P Q}{A B-P Q}=\frac{6}{12} \)
\(\frac{P Q}{12-P Q}=\frac{6}{12} \)
12 PQ = 6(12 - PQ)
12 PQ = 72- 6 PQ
12 PQ + 6 PQ = 72
18 PQ = 72
PQ = \(\frac{72}{18}=4\)
PQ = 4 cm
Since PQ = RB we have
PQ = RB = 4 cm

Construction
Step 1 : Draw a line segment QR = 5 cm.
Step 2 : At Q draw QE such that \(\angle\)RQE = 30^o.
Step 3 : At Q draw QF such that \(\angle EQF\) = 90^o
Step 4 : Draw the perpendicular bisector XY to QR which intersects QF at O and QR at G.
Step 5 : With O as centre and OQ as radius draw a circle.
Step 6: From G mark an arc in the line XY at M, such that GM = 42. cm.
Step 7 : Draw AB through M which is parallel to QR.
Step 8 : AB meets the circle at P and S.
Step 9 : Join QP and RP. Then\(\triangle\)PQR is the required triangle

Construction
Step 1: Draw a line segment PQ = 8cm.
Step 2: At P, draw PE such that \(\angle\)QPE = 60^o.
Step3: At P, draw PF such that\(\angle\)EPF = 90^o.
Step 4: Draw the perpendicular bisector to PQ, which intersects PF at O and PQ at G.
Step 5: With O as centre and OP as radius draw a circle.
Step 6: From G mark arcs of radius 5.8 cm on the circle. Mark them as R and S.
Step 7: Join PR and RQ. Then \(\triangle\)PQR is the required triangle.
Step 8: From R draw a line RN perpendicular to \(\angle\)Q. \(\angle\)Q meets RN at M
Step 9: The length of the altitude is RM = 3.8 cm.

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.

In \(\triangle\) ABC we have DE || BC.
By Thales theorem, we have \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\frac { x }{ x-2 } =\frac { x+2 }{ x-1 } \) gives x(x - 1) = (x - 2)(x + 2)
When x = 4, AD = 4, DB = x - 2, AE + x + 2 = 6, EC = x - 1 = 3 
Hence, AB = AD + DB = 4 + 2 = 6, AC = AE + EC = 6 + 3 = 9
Therefore, AB = 6, AC = 9

Let AP and CR be the vertical poles of height 6 m and 3 m respectively
Let BQ = ym 
In CRA and BQA
o
By AA criterion of similarity
\(\triangle C R A \sim \triangle B Q A\)
Their corresponding sides are Proportional
\(\frac{A C}{A B}=\frac{C R}{B Q} \)
\(\frac{A C}{A B}=\frac{3}{y} \)
\(\mathrm{AB}=\frac{A C \times y}{3}\)
In CBQ and CAP
o
\(\therefore \triangle C B Q \sim \triangle C A P \)
\(\frac{C B}{C A}=\frac{B Q}{A P} \)
\(\frac{C B}{C A}=\frac{y}{6} \)
\(B C=\frac{y \times C A}{6} \)
\((1)+(2) \Rightarrow A B+B C =\frac{A C \times y}{3}+\frac{y \times C A}{6} \)
\(A C=y \times A C\left(\frac{1}{3}+\frac{1}{6}\right) \)
\(\frac{A C}{A C} =y\left(\frac{2+1}{6}\right) \)
\(1 =y\left(\frac{3}{6}\right) \)
\(1 =\frac{1}{2} y \)
y = 2m.

Given ABC - DEF
then we have
\(\frac{\operatorname{Area}(\Delta \mathrm{ABC})}{\text { Area }(\Delta \mathrm{DEF})}=\frac{A B^{2}}{D E^{2}}=\frac{B C^{2}}{E F^{2}}=\frac{A C^{2}}{D F^{2}} \)
\(\frac{9}{16}=\frac{B C^{2}}{E F^{2}} \)
\(\frac{9}{16} =\frac{2.1 \times 2.1}{E F^{2}} \)
\(\mathrm{EF}^{2} =\left(\frac{2.1 \times 4}{3}\right)^{2} \)
\(\mathrm{EF} =\frac{2.1 \times 4}{3}=2.8 \)
EF = 2.8 cm

Let AC is the lamp post and ED is the girl.
From the triangles ABC and DBE
o
By AA criteria
Their sides are Proportional
\(\frac{A C}{D E}=\frac{B C}{B E} \)
\( \frac{A C}{12.5}=\frac{6.6}{2.5} \)
\(A C= \frac{6.6 \times 12.5}{2.5}=\frac{6.6 \times 12.5^{5}}{2.5}=33 \mathrm{~m} \)
Height of the lamp post = 33 m

Let AB and CD be two poles of height ‘a’ metres and ‘b’ metres respectively such that the poles are ‘p’ metres apart. That is AC = p metres. Suppose the lines AD and BC meet at O, such that OL = h metres
Let CL = x and LA = y.
Then, x + y = p
In \(\Delta ABC\) and \(\Delta LOC\), we have
\(\angle CAB=\angle CLO\) [each equal to 90^o]
\(\angle C=\angle C\) [ C is common]
\(\Delta CAB\sim \Delta CLO\) [By AA similarity]
\(\frac { CA }{ CL } =\frac { AB }{ LO } \) gives \(\frac { P }{ x } =\frac { a }{ h } \)
so, \(x=\frac { ph }{ a } \)   ..(1)
In \(\Delta ALO\) and \(\Delta ACD\), we have
\(\angle ALO=\angle ACD\) [each equal to 90° ]
\(\angle A=\angle A[\mathrm{~A} \text { is common }]\)
\(\frac { AL }{ AC } =\frac { OL }{ DC } \) gives \(\frac { y }{ p } =\frac { h }{ b } \) we get, \(y=\frac { ph }{ b } \)     ....(2)
(1) + (2) gives \(x+y=\frac { ph }{ a } +\frac { ph }{ b } \)
\(p=ph\left( \frac { 1 }{ a } +\frac { 1 }{ b } \right) \) (since x + y = p)
\(1=h\left( \frac { a+b }{ ab } \right) \)
Therefore, \(h=\frac { ab }{ a+b } \)
Hence, the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is \(\frac { ab }{ a+b } \) meters.

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

Given \(\frac{A D}{D B}=\frac{3}{4}\)
AC = 15 cm
EC = AC - AE
= 15 - AE

By Basic proportionality theorem
We have \(\frac{A D}{D B} =\frac{A E}{E C} \)
\(\frac{3}{4} =\frac{A E}{A C-A E} \)
\(\frac{3}{4} =\frac{A E}{15-A E} \)
3(15 - AE) = 4AE
45 - 3AE = 4AE
45 = 4AE + 3AE
7AE = 45
\(\mathrm{AE}=\frac{45}{7}=6.43 \mathrm{~cm}\)