Question

if you give me correct answer I will mark you as a brainlist pls​

Answer 1

Answer:

It is given that ∠BAD=∠EAC

It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]

It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAE

It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAEIn △BAC and △DAE

It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAEIn △BAC and △DAEAB=AD (Given)

It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAEIn △BAC and △DAEAB=AD (Given)∠BAC=∠DAE (Proved above)

It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAEIn △BAC and △DAEAB=AD (Given)∠BAC=∠DAE (Proved above)AC=AE (Given)

It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAEIn △BAC and △DAEAB=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:

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