Question

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

ar(ΔABE) = ar(ΔACF)

Answer 1

Here, BE || CY and CF || AB

$\implies \: \: \square$ EBCY is a parallelogram

《 When opposite sides of a quadrilateral are parallel, then it is a parallelogram 》

So, BE = CY --- ①

《Opposite sides of a $\sf{||^{gm}}$ are equal》

Similarly,$\square$ EBCY is a parallelogram

So, BX = CF --- ②

《Opposite sides of a $\sf{||^{gm}}$are equal》

Now, in △EBX and △YCF

$\sf{ \footnotesize{Corresponding \: Angles}} \begin{cases} \sf{∠XEB = ∠FYC} \\\sf{∠EBX = ∠YFC}\end{cases}$

$\implies$ ∠EBX = ∠YCF --- ③

[Angle Sum Property of △]

$\sf{From \: eqⁿ ①② and ③, △EBX \cong △YCF }$

$\implies$ ar(EBX) = ar(YCF) --- ④

Now, ar(AEX) = ar(AYF) --- ⑤

[Triangles having having collinear and equal bases have a common vertex A]

$\implies$ ar(AEX) + ar(EBX) = ar(AYF) + ar(YCF)

《Adding eqⁿ ④ and ⑤》

$\implies$ ar(ΔABE) = ar(ΔACF)