Question

In the given figure, AB = CF, EF = BD $\angle AFE=\angle DBC$. Prove that $\Delta AFE=\Delta CBD$

Answer 1

Given in figure

$AB=CF$ ........... (1)

$EF=BD$ ........... (2)

$\angle AFE=\angle DBC$

Then $AF=AB+BF$

And $BC=BF+CF$

As per (1) and (2) we get

$BC=BF+AB=AF$

Thus in $\Delta AFE$ and $\Delta BDC$

$BC=AF$

$EF=BD$

$\angle AFE=\angle DBC$

$\therefore \Delta AFE=\Delta CBD$ .......... [By SAS rule of congruence]