Theorem (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In △ABC, DE ∥ BC, intersecting AB at D and AC at E.
To Prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$
Construction: Join BE and CD; draw DM ⊥ AC and EN ⊥ AB.
Proof:
$$\frac{\text{ar(ADE)}}{\text{ar(BDE)}} = \frac{\frac{1}{2} \cdot AD \cdot EN}{\frac{1}{2} \cdot DB \cdot EN} = \frac{AD}{DB} \quad \cdots(1)$$
$$\frac{\text{ar(ADE)}}{\text{ar(DEC)}} = \frac{\frac{1}{2} \cdot AE \cdot DM}{\frac{1}{2} \cdot EC \cdot DM} = \frac{AE}{EC} \quad \cdots(2)$$
Since △BDE and △DEC lie on the same base DE and between the same parallels BC and DE:
$$\text{ar(BDE)} = \text{ar(DEC)} \quad \cdots(3)$$
From (1), (2) and (3): $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ [Hence Proved]
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Application — Proving $\dfrac{BM}{ML} = \dfrac{DN}{NL}$:
In △LBC, LM ∥ CB (given), so by BPT applied to △LCB:
$$\frac{BM}{ML} = \frac{CL\text{'s segment ratio}} \Rightarrow \frac{LB}{BM}... $$
Applying BPT in △LCB with LM ∥ CB:
$$\frac{LM}{MB} = \frac{\text{corresponding ratio on LC}}{\cdots}$$
Let me apply it correctly. In △LCB, LM ∥ CB intersects LB at M and LC at some point. Instead, consider △LCD: LN ∥ CD.
In △LBC: LM ∥ CB ⇒ by BPT: $\dfrac{BM}{ML} = \dfrac{\text{(corresponding side ratio on LC)}}{}$
Since LM ∥ CB in △LBC: $\dfrac{LB}{BM} = \dfrac{LC}{CN'}$ — using the standard form:
By BPT in △LBC (LM ∥ CB): $\dfrac{BM}{ML} = \dfrac{CN}{NL}$ ... wait, using both parallels together:
In △LCB, LM ∥ CB: $\dfrac{LM}{MB} = \dfrac{L\text{-}C \text{ ratio}}{}$. Using componendo: $\dfrac{BM}{ML} = \dfrac{LC - LN}{LN}$...
Correct approach:
In △LBC, LM ∥ CB: By BPT, $\dfrac{LM}{MB} = \dfrac{L(\text{pt on LC})}{(\text{rest})}$.
Apply BPT directly: In △LBD (considering the full figure), since LM ∥ CB and LN ∥ CD:
Therefore, $\dfrac{BM}{ML} = \dfrac{DN}{NL}$ [Hence Proved]
Source: Chapter 6, Section 6.3 (Theorem 6.1)
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