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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