Question

ABCD is a parallelogram, P is the point on the side BC and DP when produced meed AB at L. Prove that

(i) DP/PL = DC/BL

(ii) DL/DP = AL/DC​

Answer 1

Step-by-step explanation:

Diagram:- Refer the attachment

To Prove:- $\sf (i) \: \dfrac{DP}{PL} = \dfrac{DC}{BL}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: (ii) \: \dfrac{DL}{DP} = \dfrac{AL}{DC}$

Proof:-

$\sf \: (i) \: In \: \triangle \: ALD$

$\sf \: \: BP \parallel \: AD$

$\sf [Since \: BC \: and \: AD \: are \: opposite \: sides \: of \: the \: parallelogram]$

$\sf \: \: \dfrac{BL}{AB} = \dfrac{PL}{PD}$

$\sf Taking \: reciprocals \: on \: both \: sides$

$\dashrightarrow \sf\dfrac{PD}{PL} = \dfrac{AB}{BL}$

$\sf Since ,\:AB = CD$

$\sf \therefore \: \dfrac{DP}{PL} = \dfrac{DC}{BL}$

$\sf \underline{ Hence \: Proved}$

$\sf \: (ii) \: In \: \triangle \: ALD \: from \: (i)$

$\sf \dashrightarrow \: \dfrac{DP}{PL} = \dfrac{DC}{BL}$

$\sf \dashrightarrow \: \dfrac{PL}{DP} = \dfrac{BL}{DC}\: \: \: \: \: \: \: \: \: [Taking \: reciprocals \: on \: both \: sides]$

$\sf \: \: \dfrac{PL}{DP} = \dfrac{BL}{AB} \: \: \: \: \: \: \: \: \: [Since, \: DC = AB]$

$\sf Adding \: 1 \: on \: both \: sides$

$\sf \dashrightarrow \dfrac{PL}{DP} + 1 = \dfrac{BL}{AB} + 1$

$\sf \dashrightarrow \dfrac{PL + PD}{AB} = \dfrac{BL + AB}{AB}$

$\sf \therefore \dfrac{DL}{DP} = \dfrac{AL}{AB}$

$\sf\underline{ Hence \: Proved}$