Question

PS is the bisector of angle QPR of a ∆PQR. Prove that QS/SR = PQ/PR

Do it by indicating angle 1 angle 2 etc.

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Answer 1

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Construction:-

DRAW RX // PS to meet them to QP ...



Prove:-

in <QPS ,<SPR, <PXR , <PRX....



<SPR=<PRX (Alternate interior angle(1)



<QPS=<PXR....(Corresponding Angle(2)



then,....


From Equation (1) and (2)


<SPR=<QPS

Also------...


<PRX=<PXR

then....

PX=PR(Opposite sides of angle always Equal to Opposite Angle

Here,

In ∆PXR ....


PX=PR


Now...

In∆QRX......

=)RX //PS



Now.....
$\frac{qs}{qp} = \frac{sr}{px}$
$\frac{qs}{qp} = \frac{px}{pr}$

Note:-PX=PR (Using)




$\ll \lll \bold{ \huge{ \fbox {\color{Indigo}{It's Proved}}}} \ll \ll \lll$



$\ll \lll \bold{ \huge{ \fbox {\color{tomato}{BE BRAINLY}}}} \ll \ll \lll$


$\ll \lll \bold{ \huge{ \fbox {\color{Blue}{Yaduvanshi King}}}} \ll \ll \lll$

Answer 2

Construction:-

DRAW RX // PS to meet them to QP ...



Prove:-

in <QPS ,<SPR, <PXR , <PRX....



<SPR=<PRX (Alternate interior angle(1)



<QPS=<PXR....(Corresponding Angle(2)



then,....


From Equation (1) and (2)


<SPR=<QPS

Also------...


<PRX=<PXR

then....

PX=PR(Opposite sides of angle always Equal to Opposite Angle

Here,

In ∆PXR ....


PX=PR


Now...

In∆QRX......

=)RX //PS



Now.....
\frac{qs}{qp} = \frac{sr}{px}qpqs​=pxsr​ 
\frac{qs}{qp} = \frac{px}{pr}qpqs​=prpx​ 

Note:-PX=PR (Using)