From (i) and (ii) we get
\(\begin{array}{r} A C^{2}+A B^{2}=A D^{2}+B C \cdot D E+\frac{1}{4} B C^{2}+A D^{2}- B C \cdot D E+\frac{B C^{2}}{4} \end{array}\)
\(=2 A D^{2}+2\left(\frac{B C^{2}}{4}\right)\)
\(A C^{2}+A B^{2}=2 A D^{2}+\frac{B C^{2}}{2}\)
\(b^{2}+c^{2}=2 p^{2}+\frac{a^{2}}{2}\)

Again in \(\triangle A B C, \angle A E D=\angle A E B=90^{\circ}\)
By Pythagoras theorem.
AB² = AE² + EB²
= AD² - DE² + (BD - DE)²
= AD² - DE² + BD² + DE² -2BD.DE
= AD² + BD² - 2BD.DE
\( \mathrm{AB}^{2} =\mathrm{AD}^{2}+\left(\frac{1}{2} B C\right)^{2}-2 \cdot \frac{1}{2} \mathrm{BC} \cdot \mathrm{DE} \)
\(\mathrm{AB}^{2} =\mathrm{AD}^{2}+\frac{1}{4} B C^{2}-\mathrm{BC} \cdot \mathrm{DE} \)
\(\mathrm{AB}^{2} =\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} B C^{2} \ldots(2) \)
\(\mathrm{c}^{2} =\mathrm{p}^{2}-\mathrm{ax}+\frac{a^{2}}{4}\)  

From (1) and (2) we have
\(\frac{A Q}{A B} =\frac{A R}{A D} \)
\(\frac{A B}{A Q} =\frac{A D}{A R} \)
\(\frac{A Q+Q B}{A Q} =\frac{A R+R D}{A R} \)
\(1+\frac{Q B}{A Q} =1+\frac{R D}{A R} \)
\(\Rightarrow \frac{Q B}{A Q} =\frac{D R}{A R} \)

We have
\(\Delta \)QMO ~ \(\Delta \)RPN 
\(\frac { MQ }{ RP } =\frac { QO }{ RN } \) 
\(\frac{M Q}{Q R}=\frac{Q R}{R N}\)
[ \(\because\)OQRP is a square PR = QR and QO = QR]
QR x QR = MQ x RN
QR² = MQ x RN 

Also In \(\Delta \)QMO & \(\Delta \)RPN
\(\angle \)QMO~\(\angle \)RPN = 90^o
we have \(\Delta \) LOP ~ \(\Delta \)QMO and \(\Delta \)LOP ~ \(\Delta \)RPN
\(\Delta \)QMO ~ \(\Delta \)RPN

In \(\Delta \)LOP & \(\Delta \)PRN,
we have
\(\angle \)PLO = \(\angle\)NRP = 90°
and \(\angle\)LPO = \(\angle\)PNR (corresponding angles)
By AA criterion of similarity
\(\Delta \)LOP~\(\Delta \)RPN

From the figure, D is the mid point of BC.

We have \(\angle A E D=90^{\circ}\)
Given BC = a, AC = b, AB = c, ED = x, AD = P and AE = h
In \(\triangle\)AEC, by Pythagoras theorem
AC² =  AE² + EC²
AC² = AE² + (ED + DC)²
AC² = AE² + ED² + DC² + 2.E.D. DC
AC² = (AE² + ED²) + DC² + 2 ED . DC
AC² = AD² + DC² + 2 ED. DC
\(\mathrm{AC}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} B C\right)^{2}+2\left(\frac{1}{2} B C\right) D E\)
[D is the mid point of BC, BD,= DC]
\( A C^{2} =A D^{2}+\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} B C^{2} \ldots .(1) \)
\(\text { i.e., } b^{2} =\mathrm{P}^{2}+\mathrm{ax}+\frac{1}{4} a^{2} \)
\(\mathrm{~b}^{2} =\mathrm{p}^{2}+\mathrm{ax}+\frac{a^{2}}{4}\)

XY + CZ = 17cm
XZ + YW = 26cm
We know that diagonals if a rectangle bisect each other and the diagonals have equal length.
\(\therefore \text { Each diagonal }=\frac{26}{2}=13 \mathrm{~cm}\)
i.e., XZ = 13 cm and YW = 13 cm
Also given XY + YZ = 17 cm
Squaring on both sides (XY + YZ)² = 17²
\((\mathrm{XY})^{2}+(\mathrm{YZ})^{2}+2 \times(\mathrm{XY}) \times(\mathrm{YZ})=289\)
By Pythagoras theorem (XY)² + (YZ)² = XZ²
\(\therefore[\mathrm{XZ}]^{2}+2(\mathrm{XY}) \times(\mathrm{YZ})=289\)
132 + 2 x length x breadth = 289
2 x Area = 289 - 169
\(\text { Area }=\frac{289-169}{2}=\frac{120}{2}\)
The possible length and breadth are
(1,60) (2,30) (3,20) (4, 15), (5, 12) (6, 10).
In this pair the length and breadth should satisfy Pythagoras theorem for diagonal.
5,12 is the possible length and breadth.

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
\(\mathrm{AE}=\frac{45}{7}=6.43 \mathrm{~cm}\)

Given \(\Delta ABC\sim \Delta APQ\)
Their corresponding sides are proportional
\(\frac{A C}{A P}=\frac{C B}{P Q}=\frac{A B}{A Q} \)
\(\frac{A C}{2.8}=\frac{8}{4}=\frac{6.5}{A Q} \)
Taking
\(\frac{A C}{2.8}=\frac{8}{4} \)
\(A C=\frac{8 \times 2.8}{4}=5.6 \mathrm{~cm} \)
AC = 5.6 cm
Also taking \(\frac{8}{4} =\frac{6.5}{A Q} \)
\(\mathrm{AQ} =\frac{6.5}{8} \times 4=3.25 \mathrm{~cm} \)
AQ = 3.25 cm