Question

Two right-angled triangles ABC and ADC have the same base AC. If BC = DC, prove that AC bisects ∠BCD.

Answer 1

In ΔABC and ΔADC
∠BAC = ∠DAC     ...(90°)
BC = DC
AC = AC         ...(common)
Therefore, ΔABC ≅ ΔADC   ...(SSA criteria)
Hence, ∠BCA = ∠DCA
Thus, AC bisects ∠BCD.