Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove th

Question

Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove that

\frac{A M}{MB} = \frac{A N}{N D}

where

L M ∥ CB

and

L N ∥ C D

.

$Quadrilateral ABCD with LM parallel to CB and LN parallel to CD$

[IMAGE_1]

Accepted Answer

Step 1: Correct figure, given, to prove and construction.
Step 2: Correct proof of BPT (Basic Proportionality Theorem).
Step 3: In $	riangle ABC$, $LM \parallel CB$:
$\frac{AM}{MB} = \frac{AL}{LC}$ ... (1)
Step 4: In $	riangle ADC$, $LN \parallel CD$:
$\frac{AN}{ND} = \frac{AL}{LC}$ ... (2)
From (1) and (2), we have
$\frac{AM}{MB} = \frac{AN}{ND}$

Question

Solution