Given ABCD is a trapezium in which \(AB\parallel CD\) . Diagonals AC and BD intersect each other at O.
To prove \(\frac { AO }{ BO } =\frac { CO }{ DO } \)
Construction Draw \(EOF\parallel AB\) , also parallel to CD.

Proof In \(\triangle ACD,OE\parallel CD\)    [by construction]
\(\Rightarrow \quad \frac { AE }{ ED } =\frac { AO }{ OC } \)     ... (i)
[by basic proportionality theorem]
In \(\triangle ABD,OE\parallel BA\)                [by construction]
\(\Rightarrow \quad \frac { ED }{ AE } =\frac { OD }{ OB } \)       [by basic proportionalitytheorem]
\(\Rightarrow \quad \frac { AE }{ ED } =\frac { OB }{ OD } \)      [on taking reciprocal of the terms]  .... (ii)
From Eqs. (i) and (ii), we get
\(\frac { AO }{ OC } =\frac { OB }{ OD } \)
\(\therefore \quad \frac { AO }{ BO } =\frac { CO }{ DO } \)
Hence proved.

In \(\triangle ACB\),    \(LM\parallel CB\)   [given]
\(\Rightarrow \quad \frac { AM }{ AB } =\frac { AL }{ LC } \)    ... (i)
[by basic proportionality theorem]
In \(\triangle ACD\),     \(LN\parallel CD\)  [given]
\(\Rightarrow \quad \frac { AN }{ ND} =\frac { AL }{ LC } \) ... (ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac { AM }{ MB } =\frac { AN }{ ND } \quad \Rightarrow \frac { MB }{ AM } =\frac { ND }{ AN } \)  [reciprocal the terms]
\(\Rightarrow \quad \frac { MB }{ AM } +1=\frac { ND }{ AN } +1\)  [adding 1 on both sides]
\(\Rightarrow \quad \frac { MB+AM }{ AM } =\frac { ND+AN }{ AN } \)
\(\Rightarrow \quad \frac { AM }{ AM+MB } =\frac { AN }{ AN+ND } \) 
reciprocal the terms]
\(\therefore \quad \frac { AM }{ AB } =\frac { AN }{ AD } \)   [\(\because \)  AD = AN + ND and AB = AM + MB]

(i) \(ar\left( \triangle ABC \right) =\frac { 1 }{ 2 } CP\)  [taking AB as base]

\(ar\left( \triangle ABC \right) =\frac { 1 }{ 2 } ab\)  [taking BC as base]
\(\frac { 1 }{ 2 } cp=\frac { 1 }{ 2 } ab\)
\(\Rightarrow cp=ab\)
(ii) \(cp=ab \Rightarrow \frac { 1 }{ p } =\frac { c }{ ab } \)
\(\frac { 1 }{ { p }^{ 2 } } =\frac { { c }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \Rightarrow \frac { 1 }{ { p }^{ 2 } } =\frac { { a }^{ 2 }{ +b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\frac { 1 }{ { b }^{ 2 } } +\frac { 1 }{ { a }^{ 2 } } \)

Prove \(\triangle DBA\) and \(\triangle DCB\) are similar.
\(\frac { DB }{ DA } =\frac { DC }{ DB } \)
\(\Rightarrow { DB }^{ 2 }=DA\times DC=4\times 5\)
\(\Rightarrow { DB }=2\sqrt { 5 } \) cm
In right \(\triangle BDC\) ,
BC² = BD² + CD² = \({ \left( 2\sqrt { 5 } \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 }\)
BC = \(3\sqrt { 5 } \)
As, \(\triangle DBA\sim \triangle DCB\)
\(\Rightarrow \frac { DB }{ DC } =\frac { BA }{ BC } \)
\(\Rightarrow \frac { 2\sqrt { 5 } }{ 5 } =\frac { BA }{ 3\sqrt { 5 } } \)
BA = 6 cm
BA = 6 cm, BD = \(2\sqrt { 5 } \) cm

Prove \(\triangle DQP\) and \(\triangle DBA\) are similar.
\(\frac { QP }{ BA } =\frac { DQ }{ DB } \Rightarrow \frac { z }{ x } =\frac { DQ }{ BD } \)     ... (i)
Similar, \(\frac { z }{ y } =\frac { BQ }{ BD } \)            .... (ii)
On adding Eqs. (i) and (ii), we get
\(\frac { z }{ x } +\frac { z }{ y } =\frac { DQ }{ BD } +\frac { BQ }{ BD } =1\)
\(\Rightarrow \frac { z }{ x } +\frac { z }{ y } =1\)
\(\Rightarrow \frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 1 }{ z } \)

Prove \(\triangle ADB\sim \triangle ADC\), then \(\frac { BD }{ AD } =\frac { AD }{ CD } \)
\(\Rightarrow \)  AD² = BD.CD

In Figure (i), corresponding sides are not in proportion.
In Figure (ii), In \(\triangle LMN\)
\(\angle LMN+\angle MNL+\angle MLN\) = 180°
[by angle sum property of a triangle]
45° + \(\angle MNL\) + 57° = 180°
\(\angle MNL\) = 78°
\(\angle QPR=\angle LMN\) = 45°
\(\angle PQR = \angle MNL\) = 78°
So, \(\angle PQR\sim \angle MNL\)   [by AA similarity criterion]

Yes, 16 : 1 ; unity of nation, fraternity and patriotism.

Given, \(\triangle ABC\sim \triangle DEF\)
\(\therefore \frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle DEF \right) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \)
[ using property of area of similar triangles]
\(\Rightarrow \frac { 64 }{ 121 } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \Rightarrow \left( \frac { BC }{ EF } \right) ^{ 2 }=\left( \frac { 8 }{ 11 } \right) ^{ 2 }\)
\(\Rightarrow \frac { BC }{ EF } =\frac { 8 }{ 11 } \) [taking positive square root on both sides]
\(\Rightarrow BC=\frac { 8 }{ 11 } \times EF\)
\(\therefore BC=\frac { 8 }{ 11 } \times 15.4=11.2cm\quad \left[ \because EF=15.4cm,given \right] \)

Draw the figure according to given condition, then show that both triangles are similar by AAA similarity criterion and then use ratio of sides of both triangles to get the requried length.
10 m