Question

In the figure, if $LM \parallel CB$ and $LN \parallel CD$, then prove that $\dfrac{AM}{BM} = \dfrac{AN}{DN}$. 

Hint:

Use the Basic Proportionality Theorem (Thales’ Theorem) — if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

Answer 1

Step 1: Given.
In quadrilateral \(ABCD\), \(LM \parallel CB\) and \(LN \parallel CD\). We are to prove that: \[ \dfrac{AM}{BM} = \dfrac{AN}{DN} \]
Step 2: Apply Basic Proportionality Theorem (BPT).
From the first pair of parallel lines, \(LM \parallel CB\): \[ \dfrac{AM}{MB} = \dfrac{AL}{LC} \quad \text{(i)} \] From the second pair of parallel lines, \(LN \parallel CD\): \[ \dfrac{AN}{ND} = \dfrac{AL}{LC} \quad \text{(ii)} \]
Step 3: Compare equations (i) and (ii).
Since the right-hand sides are equal, we get \[ \dfrac{AM}{MB} = \dfrac{AN}{ND} \]
Step 4: Conclusion.
Hence proved that \[ \dfrac{AM}{BM} = \dfrac{AN}{DN} \]