In $\mathrm{â–³}CDB$,

${BC}^2={CD}^2+{BD}^2\text{Â}\text{[Pythagoras theorem]}$

${BC}^2={CD}^2+{(DA+AB)}^2$ 

${BC}^2={CD}^2+{DA}^2+{AB}^2+(2×DA×AB)\text{Â}...\left(i\right)$

In $\mathrm{â–³}ADC$,

${CD}^2+{DA}^2={AC}^2\text{Â}...\left(ii\right)\text{Â}\text{[Pythagoras Theorem]}$
Here, $\mathrm{âˆ}CAB={120}^{∘}$ (given)
$⇒\mathrm{âˆ}CAD={60}^{∘}$ (since  $\mathrm{âˆ}CAD$ and $\mathrm{âˆ}CAB$ form a linear pair of angles)

Also, $\cos {60}^{∘}=\frac{AD}{AC}$

$AC=2AD\text{Â}...\left(iii\right)$

Substituting  the values from $\left(ii\right)\text{Â}\&\text{Â}\left(iii\right)\text{Â}\text{in}\text{Â}\left(i\right)$ we get,

${BC}^2={AC}^2+{AB}^2+(AC×AB)$

$a^2=b^2+c^2+bc$

Alternatively,

 Since  $\mathrm{âˆ}A$ is an obtuse angle in  $\mathrm{â–³}ABC$ so,

${BC}^2={AB}^2+{AC}^2+2AB.AD\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}={AB}^2+{AC}^2+2×AB×\frac{1}{2}×AC\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}[âˆ\mu AD=AC\cos {60}^{∘}=\frac{1}{2}AC]\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}={AB}^2+{AC}^2+AB×AC$
 

  $⇒a^2=b^2+c^2+bc$

In $△$ABC,

DE $\parallel$ BC      (Given)

Therefore, $\frac{AD}{AB}$ = $\frac{AE}{AC}$    [By Basic Proportionality Theorem]

Comparing the above with $\frac{AD}{AB}$ = $\frac{AE}{x}$

$x$ = AC

$\angle S={180}^o-(\angle A+\angle C)$
The angles of the traingle are known. With the given measurement of the angles, we can draw infinite number of traingles (Similar triangles).

It is given that

$\frac{PS}{SQ}$ = $\frac{PT}{TR}$

So, ST||QR     [By converse of Basic Proportionality Theorem]

$∴$, $\mathrm{âˆ}$PST=$\mathrm{âˆ}$PQR (Corresponding Angles)

Also, it is given that

$\mathrm{âˆ}$PST = $\mathrm{âˆ}$PRQ

So,$\mathrm{âˆ}$PRQ =  $\mathrm{âˆ}$PQR

Therefore, PQ = PR (Sides opposite the equal angles)

i.e., $\mathrm{Δ}$PQR is an isosceles triangle.

 

Let AB be a units long.
We know that diagonal of a square = $\sqrt{2}×$ side length
$⇒AC=\sqrt{2}×AB$

Now, $\mathrm{△}$ ABE and $\mathrm{△}$ ACF are equilateral triangles.

Each angle of both $\mathrm{△}$ ABE and $\mathrm{△}$ ACF is ${60}^{∘}$.

$∴\mathrm{△}ABE∼\mathrm{△}$ ACF (by AAA similarity criterion)

So, $\frac{ar\left(ACF\right)}{ar\left(ABE\right)}=\frac{A{C}^2}{A{B}^2}=\frac{2a^2}{a^2}=\frac{2}{1}$

Consider  ΔABD and ΔACB:

 

∠BAD =  ∠BAC    [common angle]

∠BDA =  ∠ABC     [ ${90}^{∘}$]

By AA similarity criterion,
 $\mathrm{â–³}\text{Â}ABD\text{Â}\text{~}\text{Â}\mathrm{â–³}$ACB

Hence,

$\frac{ar\left(ΔABD\right)}{ar\left(ΔACB\right)}=({\frac{AB}{AC})}^2=({\frac{AD}{AB})}^2=({\frac{BD}{CB})}^2$ 

Let the triangles be $△ABC$ and $△DEF$
Then  $AB=3$ units and $DE=4$ units.

Each angle of both $△ABC$ and $△DEF$ is ${60}^{\circ }$.

$△ABE\sim △$ ACF
( AAA similarity)

So,
$\frac{ar\left(ABC\right)}{ar\left(DEF\right)}=\frac{B{C}^2}{E{F}^2}=\frac{3^2}{4^2}=\frac{9}{16}$

Consider the above figure:

AD is the part of the ladder in the first case and AC is the ladder in the second case

AB = BD = 10 $\sqrt{7}$m          [given]

2AD = AC        [since AD is the half of the ladder AC]

$∴$ By Pythagoras Theorem in  $\mathrm{△}$ABD,

${AD}^2$ =  ${AB}^2$ +  ${BD}^2$

⇒ ${AD}^2$ = $({10\sqrt{7})}^2$   +$({10\sqrt{7})}^2$ 
             = 2 $({10\sqrt{7})}^2$  

⇒$AD=10\sqrt{7}$$\sqrt{2}$

Since AD is half of the ladder, the complete length of the ladder

AC = 2AD = 20 $\sqrt{2}$$\sqrt{7}$

Now,
In  ΔABC,  
${AC}^2$ =  ${AB}^2$ +  ${BC}^2$

⇒ ${BC}^2={AC}^2−{AB}^2$
             $={\left(20\sqrt{7}\sqrt{2}\right)}^2$ -  $({10\sqrt{7})}^2$
             $=\left({10\sqrt{7})}^2\right[{\left(2\sqrt{2}\right)}^2−1]$
             = $({10\sqrt{7})}^2×7$

$∴BC=10\sqrt{7}×\sqrt{7}=70\text{Â}m$

Let $AB$ and $CD$ be two trees such that $AB=20\text{Â}m,\text{Â}CD=28\text{Â}m\text{Â}\&\text{Â}BD=17\text{Â}m$

Draw $BE$  parallel to $AC$. Then,
$ED=8\text{Â}m$.
By applying Pythagoras' theorem,
$B{E}^2+D{E}^2=B{D}^2$.

$∴BE=\sqrt{{BD}^2−{DE}^2}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}=\sqrt{{\left(17\right)}^2−8^2}$
           $=\sqrt{289−64}=\sqrt{225}=15\text{Â}m$

$∴AC=BE=15\text{Â}m$

If two triangles are similar, then the ratio of the corresponding sides are equal.
Hence,
the ratio of heights = the ratio of sides = 1 : 3.