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Thales Theorem

XY is a line ...

Question

$X Y$ is a line parallel to side $B C$ of a triangle $A B C$ . If $B E | | A C | | a n d C F | | A B$ meet $X Y$ at $E$ and $F$ respectively, Show that $a r (A B E) = a r (A C F)$ .

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Solution

$\begin{matrix} G i v e n \to Δ A B C X Y ∥ B C & B E ∥ A C & C F ∥ A B T o p r o v e \Rightarrow a r (A B E) = a r (A C F) \end{matrix}$

Proof $\to$ Let $X Y$ intersect $A B$ & $B C$ at $M$ & $N$ respectively.
$\begin{matrix} X Y ∥ B C \Rightarrow E N ∥ B C & (P a r t s o f ∥ l i n e s a r e p a r a l l e l) B E ∥ C N & (P a r t s o f ∥ l i n e s a r e p a r a l l e l) \end{matrix}$

In $B C N E,$
Both pair of opposite sides are parallel
$∵$ $B C N E$ is a parallelogram
Parallelogram $B C N E$ & $Δ$ $A E B$ lies on the same $B E$
$a r (Δ A B E) = \frac{1}{2} a r (B C N E) \to (i)$

Similarly,
$∵$ $B C F M$ is a parallelogram
$a r (Δ A C F) = \frac{1}{2} a r (B C F M) \to (i i)$

Parallelogram $B C N E$ & $B C F M$ are on same $B C$
$∵ a r (B C N E) = a r (B C F M) \to (i i i)$

From (i), (ii) and (iii), we get
$a r (A B E) = a r (A C F)$

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XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively, show that

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