Question

hello friends...

Answer this question...

Answer 1

$\bold{AE = AD -------: \: ( 1 ) \: \: [ \: Given \: ] } \: \\ \bold{ BD = CE -------: \: ( 2 ) \: \: [ \: Given \: ] }$




$\text{\underline{Subtract \: ( 2 ) \: from \: ( 1 ) , }}$


⏩ AE - BD = AD - CE


⏩ AE + CE = AD + BD


$\bold{ \Rightarrow AC = AB } \: \: \: \: \: \: \: \: \: \: \: \: \text{| \: \: \: from \: figure \: \: AE + CE = AC \: \: \: and \: \: \: AD + BD = AB } \\ \\\bold{ \Rightarrow AC = AB } \: \: \: \: \: \: - - - - - - - \: : ( \: 3\: )$




From figure,

$\bold{\angle A \: is \: common \: in \: both \: \triangle \: AEB \: and \: \triangle ADC}$





So,

$\text{AD = AE \: \: \: \: \: \: \: \: \: \: \: \: \: | \: from \: ( \: i \: )} \\ \angle\text{A} = \angle\text{A \: \: \: \: \: \: \: \: \: \: \: \: \: | \: common} \\ \text{AC = AB \: \: \: \: \: \: \: \: \: \: \: \: \: | \: from \: ( \: 3\: )}$


Hence,
$\bold{\triangle \: AEB \cong \triangle \: ADC}$ By SAS




$\boxed{ \bold{ \underline{Hence, proved. }}}$

Answer 2

Given AE = AD and BD = CE.

(1)

AE + CE = AD + BD

AC = AB   ----- (i)

In ΔAEB & ΔADC

Given AE = AD

∠EAB = ∠DAC [common angle]

AC = AB(from(i))

Then, By SAS congruence, we get

ΔAEB ≅ ADC.

Hope it helps!