PS is the bisecter of triangle …
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Posted by Subodh Katiyar, 4 years, 10 months ago
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Sia ? 4 years, 10 months ago
Given: In figure. PS is the bisector of {tex}\angle QPR{/tex} of {tex}\triangle PQR{/tex}
To prove :{tex}\frac { Q S } { S R } = \frac { P Q } { P R }{/tex}
Construction: Draw RT || SP to meet QP produced in T.
Proof: {tex}\because {/tex} RT||SP and transversal PR intersects them
{tex}\therefore \angle 1 = \angle 2{/tex} (1) ..... Alt. Int. {tex}\angle s{/tex}
{tex}\because {/tex} RT||SP and transversal QT intersects them
{tex}\therefore \angle 3 = \angle 4{/tex} (2), ..... corres. {tex}\angle s{/tex}
But {tex}\angle 1 = \angle 3{/tex} ...... Given
{tex}\therefore \angle 2 = \angle 4{/tex} ......From (1) and (2)
{tex}\therefore P T = P R{/tex} (3) ......{tex}\because {/tex} Sides opposite to equal angles of a triangle are equal
Now in {tex}\Delta \mathrm { QRT }{/tex}
PS = RT .......By construction
{tex}\therefore \frac { Q S } { S R } = \frac { P Q } { P T }{/tex} ....... By basic proportionally theorem
{tex}\Rightarrow \frac { Q S } { S R } = \frac { P Q } { P R }{/tex} ......From (3)
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