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]
\( A C^{2} =A D^{2}+\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} B C^{2} \ldots .(1) \)
\(\text { i.e., } b^{2} =\mathrm{P}^{2}+\mathrm{ax}+\frac{1}{4} a^{2} \)
\(\mathrm{~b}^{2} =\mathrm{p}^{2}+\mathrm{ax}+\frac{a^{2}}{4}\)

\(\Delta AEC\quad \Delta ADB\)
\(\angle AEC=\angle ADB={ 90 }^{ 0 }\) 
\(\angle C A E=\angle B A D\)
[common By AA similarity criteria] 
 \(\Delta AEC\sim \Delta ADB\)
(ii) Their corresponding sides are Proportional
\(\frac{C A}{A B}=\frac{C E}{D B}\)
Hence proved.

OA = 4 cm, OB = 5 cm; also OA\(\bot \)BC.
OB² = OA² + AB²
5² =  4² + AB² gives AB² = 9
Therefore AB = 3 cm
BC = 2AB hence BC = 2 x 3 = 6 cm

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.

AB = 5 cm,
AC = 10 cm,
BD  = 1.5 cm,
CD =  3.5 cm.

\( \frac{A B}{A C}=\frac{5}{10}=\frac{1}{2} \)
\(\frac{B D}{D C}=\frac{1.5}{3.5}=\frac{3}{7} \)
\(\frac{A B}{A C} \neq \frac{B D}{D C} \)
By the converse of the angle bisector theorem, AD is not a bisector of \(\angle\)A

Let BD =  x cm, then DC = (6 – x)cm
AD is bisector of\(\angle\) A
Therefore by Angle Bisector Theorem
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac { 10 }{ 14 } =\frac { x }{ 6-x } \quad \frac { 5 }{ 7 } =\frac { x }{ 6-x } \)
So, 12x = 30 we get, \(x=\frac { 30 }{ 12 } =2.5\)
Therefore, BD = 2.5 cm, DC = 6−x = 6−2.5 = 3.5 cm

Given o = \(\triangle\)ABC and \(\triangle\)ADE
By AA criteria for similarity
\(\triangle A B C \sim \triangle A D E\)
Their corresponding sides are proportional
\(\therefore \frac{B C}{D E} =\frac{A C}{A E}=\frac{A B}{A D} \)
\(\frac{12}{D E} =\frac{3+2}{A E}=\frac{A B}{3} \)
Also in right ABC , o
using Pythagoras theorem, we have
AB² = BC² + AC² = 12² + 5²
= 144 + 25 = 169
AB = 13 cm
Now from (1), we get
\(\frac{12}{D E}=\frac{13}{3} \)
\(\mathrm{DE}=\frac{12 \times 3}{13}=\frac{36}{13} \)
DE = 2.77 cm
Also from (1) \(\frac{5}{A E}=\frac{13}{3} \)
\(\mathrm{AE}=\frac{5 \times 3}{13}=\frac{15}{13}=1.15 \mathrm{~cm} \)

Let DE be the vertical stick and AB is the tower,
DE = 6 m, EF = 400 cm = 4 m, BC = 28 m
From DFE and ACB
Using similarity criteria
\(\frac{A B}{D E}=\frac{B C}{E F}\)
\(\frac{A B}{6}=\frac{28}{4} \)
\(A B=\frac{28 \times 6}{4}=42 \mathrm{~m} \)
Height of the tower = 42 m

(i) In ABC and ADE <A is common
\( \frac{A E}{E C}=\frac{2}{3 \frac{1}{2}}=\frac{\frac{2}{7}}{2}=\frac{2 \times 2}{7}=\frac{4}{7} \)
\(\frac{A D}{D B}=\frac{3}{5} \)
\(Here\ \frac{4}{7} \pm \frac{3}{5} \)
\(\frac{A E}{E C} \neq \frac{A D}{D B} \)
The corresponding sides are not proportional.
ABC and ADE are not similar
(ii) In CPQ and CAB <C is common
<PQC = 180^o - 110^o = 70^o
[ <PQC and <PQB are liner pair of angles]
<ABC = 70^o
<BAC = <QPC
[ sum of three angles of a triangle are 180]
<PCQ = 180^o - (< QPC + 70^o)
<ABC = <PQC = 70^o
<C common and <BAC = <QPC
By AAA similarity criteria, <ABC <PQC
Corresponding sides are Proportional
\(\frac{A B}{P Q} =\frac{B C}{Q C} \)
\(\frac{5}{\sqrt{x}} =\frac{6}{3} \)
[BC = BQ + QC = 3 + 3 = 6]
\(x=\frac{5}{6} \times 3=2.5\)
x = 2.5

In \(\triangle\)PST and \(\triangle\)PQR,
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+1 } =\frac { 2 }{ 3 } ,\frac { PT }{ PR } =\frac { 4 }{ 4+2 } =\frac { 2 }{ 3 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is common
Therefore, by SAS similarity,
\(\triangle\) PST~\(\triangle\)PQR