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T.C., Dls cyuISLA I UL ile L UI 24.1 IUVCU 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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