In figure ,AC=AEAB=ADand traingle BAD= traingle EAC.show tat BC=DE
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AB=AD
AC=AE
angle BAD+DAC=angle EAC+DAC
so,BAC=DAC
hence,∆BAC=~DAE (by SAS property)
so,BC=DE(by cpct)
AC=AE
angle BAD+DAC=angle EAC+DAC
so,BAC=DAC
hence,∆BAC=~DAE (by SAS property)
so,BC=DE(by cpct)
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hope this will help you
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