Question

In the figure given alongside, prove that
$\left(i\right)AB=FC$
$\left(ii\right)AF=BC$
$[Hint.$ $\left(i\right)\Delta ABE\cong \Delta CFD,RHS$
$\left(ii\right)AB=FC\Rightarrow AF+FB=FB+BC.]$

Answer 1

Given

9n triangle $\Delta ABE$ and $\Delta CFD$

(i) $\angle ABE=\angle CFD$ (RIght angled)

$BE=FD$ {GIven}

$AE=DC$

$\therefore$ from RHS rule

$\Delta ABE\cong \Delta CFD$

$\therefore AB=FC$ (from congruency Rule)

(ii) 9n the above part we have proved that

$AB=FC\to AB=AF+FB$

$FC=FB+BC$

$\therefore AF+FB=FB+BC$

$\therefore AF=BC$ Hence proved