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.

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

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 a triangle \(\triangle\)PQR. We have to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the given \(\triangle\)PQR.

Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex P.
3. Joined Q₃ to R and drawn a line through Q₇ parallel to Q₃R, intersecting the extended line segment QR at R'
4. Drawn a line through R' parallel to RP intersecting the extended line segment QP at P.
5. Then PQR' is the required triangle each of whose sides is seven-thirds of the corresponding sides of PQR.

Given a triangle ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ABC

Steps of construction:
1. Constructed a  ABC with any measurement
2. Drawn a ray BX making an acute angle with BC on the side opposite to the vertex A.
3. Joined B₅ to C and drawn a line through B₆ parallel to B₅C intersecting the extended line segment BC at C.
4. Drawn a line through C' parallel to CA intersecting the extended BA at A'.
5. Then A'BC' is the required triangle each of whose sides is six-fifths of the corresponding sides of ABC.

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 3 }{ 5 } \) of the corresponding sides of the triangle PQR

 
Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex P.
Located 3 points Q₁, Q₂ and Q₃ on QX so that Q Q₁ = Q₁ Q₂ = Q₂ Q₃
4. Joined Q₃R and drawn a line through Q₂ parallel to Q₃R to intersect QR at R'.
5. Drawn a line through R' parallel to the line RP to intersect QP at P'. Then PQR is the required triangle each of whose sides is two-thirds of the corresponding sides of PQR.

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 7 }{ 4 } \) of the corresponding sides of the triangle PQR.
Steps of construction
1. Construct a DPQR with any measurement.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 4 in \(\frac { 7 }{ 4 } \))
Q_1,Q_2,Q_3,Q_4,Q_5,Q₆ and Q₇ on QX so that 
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇
4. Join Q₄ (the 4th point, 4 being smaller of 4 and 7 in \(\frac { 7 }{ 4 } \)) to R and draw a line through Q₇ parallel to Q₄R, intersecting the extended line segment QR at R'.
5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'.
Then \(\triangle\)P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of \(\triangle\)PQR.

Given a triangle PQR we are required to construct another triangle whose sides are \(\frac{3}{5}\) of the corresponding sides of the triangle PQR.



Steps of construction
1. Construct a \(\triangle\) PQR with any measurement
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 5 (the greater of 3 and 5 in \(\frac { 3 }{ 5 } \)) points.
Q₁Q_2, Q₃, Q₄ and Q₅ on QX so that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅
4. Join Q₅R  and draw a line through Q₃ (the third point, 3 being smaller of 3 and 5 in \(\frac { 3 }{ 5 } \)) parallel to Q₅R to intersect QR at R'.
5. Draw line through R' parallel to the line RP to intersect QP at P'.
Then, \(\triangle\)P'QR' is the required triangle each of whose sides is three-fifths of the corresponding sides of \(\triangle\) PQR.