Question

AC=AE , AB=AD and
angleBAD = angle EAC, show that BC=DE​

Answer 1

Step-by-step explanation:

$\color{blue}{given \: that}...$

• AC = AE
• AB = AD
• angle BAD = angle EAC

$\color{blue}{to \: prove}...$

• BC = DE

$\color{blue}{solution}...$

angle BAD = angle EAC

=> angleBAD + angleDAC = angleEAC + angleDAC

[by adding angleDAC on both sides]

=> angle BAC = angle EAD ......eq.(i)

Now in Δ BAC & Δ EAD :

AB = AD (given)

AC = AE (given)

angle BAC = angle DAE (using eq.(i)

Therefore, Δ BAC is congruent to Δ EAD

( by SAS congruence rule )

Therefore, BC = DE ( by CPCT)

$\color{red}{hence \: proved}$

I Hope it helps you.......

Answer 2

Answer:

$\large{\underline{\underline{\red{\bf{mark \: the \: above \: ans \: brainlist}}}}}$