Prerequisites:
Basic Proportionality Theorem (proof)
Unique parallel through a point (proof)
proof:
Let there be a \triangle ABC and a line l intersecting the sides AB and AC at the points D and E respectively, as shown in the figure, such that:
\qquad\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}
We need to show that l \parallel BC. For this, let us assume that l \not\parallel BC. Then there must exist a unique line DF through the point D, such that DF \parallel BC.
Since DF \parallel BC, hence by Basic Proportionality Theorem, we get:
\qquad\quad\;\:\!\dfrac{AD}{DB} = \dfrac{AF}{FC}\\[12pt] \text{But, }\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}\qquad\qquad\qquad\qquad\text{(given)}\\[12pt] \therefore\quad\quad\:\dfrac{AF}{FC} = \dfrac{AE}{EC}\\[12pt] \Rightarrow\quad\;\;\;\dfrac{AF}{FC} +1 = \dfrac{AE}{EC} +1\\[12pt] \Rightarrow\quad\;\;\;\dfrac{AF+FC}{FC} = \dfrac{AE+EC}{EC}\\[12pt] \Rightarrow\quad\;\;\;\:\dfrac{AC}{FC} = \dfrac{AC}{EC}\\[12pt] \Rightarrow\qquad\; FC = EC
But this is a contradiction as F \neq E. Hence, our assumption was false. Therefore, l \parallel BC.
Q.E.D.
Recommended:
Angle Bisector Theorem
Midpoint Theorem
Location of centroid
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