Question

Q.6) In the given figure,
AC= AE, AB = and
< BAD = < EAC. show that BC = DE​

Answer 1

Answer:

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:

Step-by-step explanation: It is given that ∠BAD=∠EAC

Step-by-step explanation: It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]

Step-by-step explanation: It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAE

Step-by-step explanation: It is given that ∠BAD=∠EAC∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides]∴∠BAC=∠DAEIn △BAC and △DAE

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

Step-by-step explanation: 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)

Step-by-step explanation: 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)

Step-by-step explanation: 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)

Step-by-step explanation: 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)