Proof: As, \(\displaystyle\overline{{{S}{P}}}\) is altitude to \(\displaystyle\overline{{{N}{R}}}\),

\(\displaystyle\Rightarrow\angle{S}{P}{N}={90}^{{\circ}}\)

Similarity, as \(\displaystyle\overline{{{R}{T}}}\) is altitude to \(\displaystyle\overline{{{N}{S}}}\),

\(\displaystyle\Rightarrow\angle{R}{T}{N}={90}^{{\circ}}\)

\(\displaystyle\Rightarrow\angle{R}{T}{N}\stackrel{\sim}{=}\angle{S}{P}{N}\)

Now, In \(\displaystyle\triangle{N}{R}{T}\ {\quad\text{and}\quad}\ \triangle{N}{S}{P}\)

\(\displaystyle\angle{R}{T}{N}\stackrel{\sim}{=}\angle{S}{P}{N}\) (Proved above)

\(\displaystyle\angle{N}\stackrel{\sim}{=}\angle{N}\) (Common)

\(\displaystyle\Rightarrow\triangle{N}{R}{T}\sim\triangle{N}{S}{P}\) (By AA Similarity Rule)

\(\displaystyle\Rightarrow\angle{S}{P}{N}={90}^{{\circ}}\)

Similarity, as \(\displaystyle\overline{{{R}{T}}}\) is altitude to \(\displaystyle\overline{{{N}{S}}}\),

\(\displaystyle\Rightarrow\angle{R}{T}{N}={90}^{{\circ}}\)

\(\displaystyle\Rightarrow\angle{R}{T}{N}\stackrel{\sim}{=}\angle{S}{P}{N}\)

Now, In \(\displaystyle\triangle{N}{R}{T}\ {\quad\text{and}\quad}\ \triangle{N}{S}{P}\)

\(\displaystyle\angle{R}{T}{N}\stackrel{\sim}{=}\angle{S}{P}{N}\) (Proved above)

\(\displaystyle\angle{N}\stackrel{\sim}{=}\angle{N}\) (Common)

\(\displaystyle\Rightarrow\triangle{N}{R}{T}\sim\triangle{N}{S}{P}\) (By AA Similarity Rule)