Question

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Answer 1

Answer:

BC=DE (By CPCT

Step-by-step explanation:

It is given that ∠BAD=∠EAC

∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]

∴∠BAC=∠DAE

In △BAC and △DAE

AB=AD (Given)

∠BAC=∠DAE (Proved above)

AC=AE (Given)

∴△BAC≅△DAE (By SAS congruence rule)

∴BC=DE (By CPCT)

Answer 2

Step-by-step explanation:

$\large \tt{it \: is \: given \: that} \angle \: BAD= \angle \: EAC$

$\large \sf{ \angle \: BAD + \angle \: DAC = \angle \: EAC + \angle \: DAC \: } \ \\ \large \sf [add \angle \: DAC \: on \: both \: sides]$

$\large \bf{ \therefore \:\angle \: BAC = \angle \: DAE}$

$\large \tt{In \angle \: BAC \: and \angle \: DAE}$

$\large \sf{AB = AD \: (Given)}$

$\large \cal{</p><p> \angle \: BAC = \angle \: DAE \: (Proved \: above)}$

$\large \mathbb{AC = AE \: (Given)}$

$\large \sf{ \therefore \: ∆BAC \cong∆DAE (by \: SAS \: conguerence \: rule)}</p><p></p><p>$

$\huge \sf{BC=DE (by \: CPCT)}$

hope it helps you ✌️