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Question
Two right-angled triangles ABC and ADC have the same base AC. If BC = DC, prove that AC bisects ∠BCD.
Solution
In ΔABC and ΔADC
∠BAC = ∠DAC ...(90°)
BC = DC
AC = AC ...(common)
Therefore, ΔABC ≅ ΔADC ...(SSA criteria)
Hence, ∠BCA = ∠DCA
Thus, AC bisects ∠BCD.
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