Question

pls help asap to solve with reasoning​

Answer 1

GIVEN:-

• AC = AE , AB = AD.

• $\rm{\angle{BAD} = \angle{EAC} }$.

TO PROVE:-

• BC = DE.

Now,

$\implies\rm{\angle{BAD} = \angle{EAC}}$.

• $\rm{\angle{DAC}}$ is a common, Adding it on Both sides.

$\implies\rm{\angle{BAD} + \angle{DAC} = \angle{EAC} + \angle{DAC}}$

$\implies\rm{\angle{BAC} = {EAD}}$......1.

Now, In ∆ABC & ∆ADE.

$\implies\rm{AB = AD}$(Given).

$\implies\rm{AC = AE}$(Given).

$\implies\rm{\angle{BAC} = \angle{EAD}}$.(From 1.).

Hence, $\rm{\angle{ABC} ≈ \angle{ADE}}$ By S - A -S criteria.

So, BC = DE by CPCT.

EXTRA INFORMATION:-

• S - S - S = When three side are equal to other corresponding triangle.

• A - S - A -= When two angles and one Side is equal to other corresponding triangle.

Answer 2

Answer:

➡GIVEN:-

✏AC = AE , AB = AD.

✏∠BAD=∠EAC .

➡TO PROVE:-

✏BC = DE.

Now,

=>∠BAD=∠EAC .

⚫∠DAC is a common, Adding it on both sides.

=>∠BAD+∠DAC=∠EAC+∠DAC

=>∠BAC=EAD ......1.

⭐Now, In ∆ABC & ∆ADE.

➡ Given:-

✏AB=AD

✏AC=AE

=>∠BAC=∠EAD .(From 1.).

Hence, ∠ABC≈∠ADE

(By S - A -S criteria.)

So, BC = DE by CPCT.

Step-by-step explanation:

HOPE IT HELP YOU