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Question
In the given figure ray AZ bisects ∠BAD and ∠DCB, prove that ∆BAC ≅ ∆DAC
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Solution
In ∆BAC and ∆DAC
∠BAC = ∠DAC ...[Given `bar("AZ")` bisects ∠BAD]
∠BCA = ∠DCA ...[`bar("AZ")` bisects ∠DCB]
AC = AC ...[∵ common side]
∴ Here AC is the included side of the angles
By ASA criterior, ∆BAC ≅ ∆DAC.
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