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Question
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Q.6. In the figure, AC = AE, AB = AD and
BAD = 2 EAC. Show that BC = DE.
Sol. ZBAD = ZEAC [Given]
> BAD + DAC = LEAC + DAC
[Adding DAC to both sides
ZBAC = ZEAC ... (i)
Now, in AABC and AADE, we have
AB = AD [Given)
AC = AE (Given)
ZBAC = ZDAE [From (i)]
AABC = AADE [By SAS congruencel
BC = DE
ICPCT) Proved.
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