To prove : The similarity of \triangle NRT with respect to \triangle NS

alesterp 2021-11-02 Answered
To prove : The similarity of \(\displaystyle\triangle{N}{R}{T}\) with respect to \(\displaystyle\triangle{N}{S}{P}\).
Given information: Here, we have given that \(\displaystyle\overline{{{S}{P}}}\) is altitude to \(\displaystyle\overline{{{N}{R}}}\ {\quad\text{and}\quad}\ \overline{{{R}{T}}}\) is altitude to \(\displaystyle\overline{{{N}{S}}}\).

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Expert Answer

Margot Mill
Answered 2021-11-03 Author has 9592 answers
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)
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