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

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
\(=20 \sqrt{13} \mathrm{~km}\)

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.

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

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