ABCD is a parallelogram. P is a point on side BC and DP when

Q) ABCD is a parallelogram. P is a point on side BC and DP when produced meets AB produced at L. Prove that:

(i) $\frac{DP}{PL} = \frac{DC}{BL}$

(ii) $\frac{DL}{DP} = \frac{AL}{DC}$

(iii) If LP : PD = 2 : 3, then find BP : BC

Ans:

(i) Since DC ǁ AB (and AL)

Line CB cuts these parallel lines,

Therefore, ∠ DCB = ∠ CBL

∠ DCP = ∠ PBL

Line DL cuts the parallel lines CD and AL,

Therefore, ∠ CDL = ∠ ALD

∠ CDP = ∠ BLP

Δ DCP ~ Δ PBL

$\therefore \frac{DP}{PL} = \frac{DC}{BL}$

(ii) Line DL cuts the parallel lines CD and AL,

Therefore, ∠ CDL = ∠ ALD

∠ CDP = ∠ ALD (interior angles)

Since ABCD is a parallelogram, therefore

Therefore, ∠ DAB = ∠ DCP (Opposite angles)

Δ DAL ~ Δ DCP

$\frac{DL}{DP} = \frac{AL}{DC}$

(iii) Since Δ DCP ~ Δ PBL

$\therefore \frac{LP}{PD} = \frac{BP}{PC}$

Given that = $\frac{LP}{PD} = \frac{2}{3}$

$\therefore \frac{BP}{PC} = \frac{2}{3}$

or PC = $\frac{3}{2}$ BP

Since BC = BP + $\frac{3}{2}$ BP

BC = $\frac{5}{2}$ BP

$\frac{BP}{BC} = \frac{2}{5}$

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