In the figure given above, AC is the diameter. AB, BC, AD, DC are chords of the circle.
AD = BC = 3 cm.
Then, ΔABC≅ΔADC
True
As the angles made by the diameter at the circumference is 90∘, so ΔABC and ΔADC are right angle triangles, right angled at B and D respectively.
In the figure given above, consider
In ΔABCIn ΔADCAB2+BC2=AC2AD2+DC2=AC2AB2+32=5232+DC2=52AB2+25−99+C2=25AB2=16DC2=25−9AB2=42DC2=16AB=4DC=4
In ΔABC and ΔADC
BC = AD (given)
AB = DC (AB = DC = 5 cm)
AC = AC (common side)
∴ΔABC≅ADC (by SSS criterion of congruency)