Conclusion:

By A-A similarity, it can be proved that:

\(\displaystyle\triangle{L}{M}{K}\sim\triangle{M}{N}{K}\)

Similar triangles have corresponding sides as proportional.

\(\displaystyle\frac{{{M}{K}}}{{{L}{K}}}=\frac{{{N}{K}}}{{{M}{K}}}\)

\(\displaystyle\Rightarrow\frac{{{3}{K}{P}}}{{25}}=\frac{{9}}{{{3}{K}{P}}}\)

\(\displaystyle\Rightarrow{K}{P}={5}\)

\(\displaystyle\Rightarrow{K}{M}={3}{K}{P}={15}\)

By A-A similarity, it can be proved that:

\(\displaystyle\triangle{L}{M}{K}\sim\triangle{R}{M}{P}\)

Similar triangles have corresponding sides as proportional.

\(\displaystyle\frac{{{M}{K}}}{{{L}{K}}}=\frac{{{P}{M}}}{{{R}{P}}}\)

\(\displaystyle\Rightarrow\frac{{15}}{{25}}=\frac{{10}}{{{R}{P}}}\)

\(\displaystyle\Rightarrow{K}{P}={16}\frac{{2}}{{3}}\)

Using Pythagorean Theorem in right angled triangle \(\displaystyle\triangle{L}{M}{K}\):

\(\displaystyle{H}{y}{p}{o}{t}{e}\nu{s}{e}^{{2}}={B}{a}{s}{e}^{{2}}+{P}{e}{r}{p}{e}{n}{d}{i}{c}\underline{{a}}{r}^{{2}}\)

\(\displaystyle={25}^{{2}}={15}^{{2}}+{L}{M}^{{2}}\)

\(\displaystyle\Rightarrow{L}{M}^{{2}}={400}\)

\(\displaystyle\Rightarrow{L}{M}={20}\)

In similar triangles: \(\displaystyle\triangle{L}{M}{K}\sim\triangle{M}{N}{K}\)

Ratio can be written as:

\(\displaystyle\frac{{{L}{M}}}{{{M}{K}}}=\frac{{{M}{N}}}{{{N}{K}}}\)

\(\displaystyle\Rightarrow\frac{{20}}{{15}}=\frac{{{M}{N}}}{{9}}\)

\(\displaystyle\Rightarrow{M}{N}={12}\)

\(\displaystyle{K}{P}={5},{K}{M}={15},{M}{R}={13}\frac{{1}}{{3}},{M}{L}={20},{M}{N}={12},{P}{R}={16}\frac{{2}}{{3}}\)

By A-A similarity, it can be proved that:

\(\displaystyle\triangle{L}{M}{K}\sim\triangle{M}{N}{K}\)

Similar triangles have corresponding sides as proportional.

\(\displaystyle\frac{{{M}{K}}}{{{L}{K}}}=\frac{{{N}{K}}}{{{M}{K}}}\)

\(\displaystyle\Rightarrow\frac{{{3}{K}{P}}}{{25}}=\frac{{9}}{{{3}{K}{P}}}\)

\(\displaystyle\Rightarrow{K}{P}={5}\)

\(\displaystyle\Rightarrow{K}{M}={3}{K}{P}={15}\)

By A-A similarity, it can be proved that:

\(\displaystyle\triangle{L}{M}{K}\sim\triangle{R}{M}{P}\)

Similar triangles have corresponding sides as proportional.

\(\displaystyle\frac{{{M}{K}}}{{{L}{K}}}=\frac{{{P}{M}}}{{{R}{P}}}\)

\(\displaystyle\Rightarrow\frac{{15}}{{25}}=\frac{{10}}{{{R}{P}}}\)

\(\displaystyle\Rightarrow{K}{P}={16}\frac{{2}}{{3}}\)

Using Pythagorean Theorem in right angled triangle \(\displaystyle\triangle{L}{M}{K}\):

\(\displaystyle{H}{y}{p}{o}{t}{e}\nu{s}{e}^{{2}}={B}{a}{s}{e}^{{2}}+{P}{e}{r}{p}{e}{n}{d}{i}{c}\underline{{a}}{r}^{{2}}\)

\(\displaystyle={25}^{{2}}={15}^{{2}}+{L}{M}^{{2}}\)

\(\displaystyle\Rightarrow{L}{M}^{{2}}={400}\)

\(\displaystyle\Rightarrow{L}{M}={20}\)

In similar triangles: \(\displaystyle\triangle{L}{M}{K}\sim\triangle{M}{N}{K}\)

Ratio can be written as:

\(\displaystyle\frac{{{L}{M}}}{{{M}{K}}}=\frac{{{M}{N}}}{{{N}{K}}}\)

\(\displaystyle\Rightarrow\frac{{20}}{{15}}=\frac{{{M}{N}}}{{9}}\)

\(\displaystyle\Rightarrow{M}{N}={12}\)

\(\displaystyle{K}{P}={5},{K}{M}={15},{M}{R}={13}\frac{{1}}{{3}},{M}{L}={20},{M}{N}={12},{P}{R}={16}\frac{{2}}{{3}}\)