A parallelogram ABCD in which p …

Given A parallelogram ABCD in which P is a point on side BC such that DP produced meets AB produced at L.
To Prove

1. {tex}\frac { D P } { P L } = \frac { D C } { B L }{/tex}  
2. {tex}\frac { D L } { D P } = \frac { A L } { D C }{/tex}

Proof

1. In {tex}\triangle{/tex} ALD, we have
   {tex}B P \| A D{/tex}
   {tex}\therefore \quad \frac { L B } { B A } = \frac { L P } { P D }{/tex}
   {tex}\Rightarrow \quad \frac { B L } { A B } = \frac { P L } { D P }{/tex}
   {tex}\Rightarrow \quad \frac { B L } { D C } = \frac { P L } { D P }{/tex} [{tex}\because{/tex} AB = DC]
   {tex}\Rightarrow \quad \frac { D P } { P L } = \frac { D C } { B L }{/tex} [Taking reciprocals of both sides].
2. From (i), we have
   {tex}\frac { D P } { P L } = \frac { D C } { B L }{/tex}
   {tex}\Rightarrow \quad \frac { P L } { D P } = \frac { B L } { D C }{/tex} 
   {tex}\Rightarrow \quad \frac { P L } { D P } = \frac { B L } { A B }{/tex} [{tex}\because{/tex} DC = AB]  
   {tex}\Rightarrow \quad \frac { P L } { D P } + 1 = \frac { B L } { A B } + 1{/tex} [Adding 1 on both sides]
   {tex}\Rightarrow \quad \frac { D P + P L } { D P } = \frac { B L + A B } { A B }{/tex}
   {tex}\Rightarrow \quad \frac { D L } { D P } = \frac { A L } { A B }{/tex}
   {tex}\Rightarrow \quad \frac { D L } { D P } = \frac { A L } { D C }{/tex} [ {tex}\because{/tex} AB = DC]​​​​​