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.

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

BL and CM are medians at the \(\triangle\)ABC in which
\(A=\angle { 90 }^{ 0 }\)
From \(\triangle\)ABC
BC² = AB² + AC²
(Pythagoras theorem)

From \(\Delta ABL\)
BL² = AL² + AB²
\({ BL }^{ 2 }=\left( \cfrac { { AC }^{ 2 } }{ 2 } \right) +{ AB }^{ 2 }\)
(L is the mid-point at AC)
\({ BL }^{ 2 }=\cfrac { { AC }^{ 2 } }{ 4 } +{ AB }^{ 2 }\)
4BL² = AC² + 4AB²
From \(\Delta CMA\)
CM² = AC² + AM²
\({ CM }^{ 2 }={ Ac }^{ 2 }+\left( \cfrac { AB }{ 2 } \right) ^{ 2 }\)
(M is the mid-point at AB)
\({ CM }^{ 2 }={ AC }^{ 2 }+\cfrac { { AB }^{ 2 } }{ 4 } \)
4CM² = 4AC²+ AB²
Adding (2) and (3), we have
4(BL² + CM²) = 5(AC² + AB²)
4(BL² + CM²) = 5BC²

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

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

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

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.

Construction
Step 1 : Draw a line segment BC = 8 cm.
Step 2 : At B, draw BE such that \(\angle\)CBE = 60^o.
Step 3 : At B, draw BF such that \(\angle\)EBF = 90^o .
Step 4 : Draw the perpendicular bisector to BC, which intersects BF at O and BC at G.
Step 5 : With O as centre and OB as radius draw a circle.
Step 6 : From B, mark an arc of 6 cm on BC at D.
Step 7 : The perpendicular bisector intersects the circle at I. Joint ID.
Step 8 : ID produced meets the circle at A. Now join AB and AC. Then \(\triangle\)ABC is the required triangle.