Since, each of the \(\triangle BCE\) and \(\triangle ACF\) is an equilateral triangle, so each angle one of them is 60°.

\(\triangle BCE\) and \(\triangle ACF\) are equiangular
\(\triangle BCE\sim \triangle ACF\)
Since, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding side.
\(\frac { ar\left( \triangle BCE \right) }{ ar\left( \triangle ACF \right) } =\frac { { BC }^{ 2 } }{ { AC }^{ 2 } } =\frac { { BC }^{ 2 } }{ 2{ \left( BC \right) }^{ 2 } } =\frac { 1 }{ 2 } \) \(\left[ \because AC=\sqrt { 2BC } \right] \)

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

Prove \(\triangle EDC\) and \(\triangle EBA\) are similar.
Then, \(\frac { ED }{ EB } =\frac { EC }{ EA } \)
\(\frac { ED }{ EC } =\frac { EB }{ EA } \)       ... (i)  

Now, prove \(\triangle AED\) and \(\triangle BEC\) are similar
\(\frac { ED }{ EC } =\frac { EA }{ EB } =\frac { AD }{ BC } \)          ...(ii)
From Eqs.(i) and (ii), \(\frac { EB }{ EA } =\frac { EA }{ EB } \)
EB² = EA² \(\Rightarrow \)  EB = EA
From Eq.(ii), we get
\(\frac { EA }{ EA } =\frac { AO }{ BC } \Rightarrow 1=\frac { AD }{ BC } \Rightarrow AD=BC\)

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

As \(\triangle ABC\sim \triangle DFE\)
\(\angle D=\angle A={ 30 }^{ ° }\)
\(\angle C=\angle E={ 50 }^{ ° }\)
\(\angle B=\angle F={ 180 }^{ ° }-\left( { 50 }^{ ° }+{ 30 }^{ ° } \right) ={ 100 }^{ ° }\)
Now, \(\frac { AB }{ DF } =\frac { AC }{ DE } \)
\(\frac { 5 }{ 7.5 } =\frac { 8 }{ DE } \)
\(\Rightarrow DE=12cm,\angle F={ 100 }^{ ° }\)

100 cm²

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \)
\(\frac{P F}{F R}=\frac{3.6}{2.4}=1.5\)
Hence, \(\frac{P E}{E Q} \neq \frac{P F}{F R}\)
Therefore, EF is not parallel to QR

9 : 7

Find BC using Pythagoras theorem and then find AC.
AC = \(\sqrt{369}\) m

Prove \(\triangle ABC\) and \(\triangle AMP\)  are similar.
then take ratio \(\frac{AC}{AP}=\frac{BC}{MP}\)

Let AB be the vertical pole of height 18 m and guy wire BC of length 24 m. Let AC = x m be the distance of the stake from the base of the pole.
In \(\triangle BAC\), using Pythagoras theorem, we get
\({ AC }^{ 2 }+{ AB }^{ 2 }={ BC }^{ 2 }\Rightarrow { x }^{ 2 }+{ \left( 18 \right) }^{ 2 }={ \left( 24 \right) }^{ 2 }\)
[ \(\because \) AC = x m, AB = 18 m and BC = 24 m]
\(\Rightarrow { x }^{ 2 }={ \left( 24 \right) }^{ 2 }-{ \left( 18 \right) }^{ 2 }=576-324=252\)
\(\therefore x=\sqrt { 252 } =6\sqrt { 7 } \)
[taking positive square root]
Hence, the distance of the stake from the base of pole is \(6\sqrt { 7 } \) m.

Diagonals of a rhombus bisect each other at right angles.

So, OA = OC = 8 cm
and OB = OD = 6 cm
Now use pythagoras theorem in  ΔAOB
10 cm.