If angle 1 = angle 2 …
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Preeti Dabral 1 year, 9 months ago
Since {tex}\triangle {/tex} NSQ {tex}\cong{/tex} {tex}\triangle {/tex}MTR
{tex}\therefore {/tex} {tex}\angle{/tex} SQN = {tex}\angle{/tex} TRM
{tex}\Rightarrow {/tex} {tex}\angle{/tex} Q= {tex}\angle{/tex} R(in {tex}\triangle {/tex}PQR)
{tex}\angle Q{/tex} = 90o - {tex}\frac{1}{2}\angle P{/tex}
Again 1 = {tex}\angle{/tex} 2 [given in {tex}\triangle {/tex} PST](Isosceles property)
{tex}\therefore {/tex} {tex}\angle{/tex} 1= {tex}\angle{/tex} 2 = {tex}\frac{1}{2}\left( {{{180}^ \circ } - \angle P} \right){/tex}
{tex}= {90^ \circ } - \frac{1}{2}\angle P{/tex}
Thus, in {tex}\triangle {/tex}PTS and {tex}\triangle {/tex}PRQ
{tex}\angle{/tex} 1 = {tex}\angle{/tex} Q [Each = 90° - {tex}\frac{1}{2}\angle P{/tex}]
{tex}\angle{/tex} 2 = {tex}\angle{/tex} R, {tex}\angle{/tex} P= {tex}\angle{/tex} P(Common)
{tex}\triangle {/tex} PTS {tex}\cong{/tex} {tex}\triangle {/tex}PRQ
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