Question

In the given figure, if $\angle$ACB = ${30}^{\circ }$, then the value of $\angle$ACD is ___${}^{\circ }$.

Answer 1

In the given figure,
AD=AB (Given)
DC=BC (Given)
AC is common
$\therefore \Delta ADC\cong \Delta ABC$ (SSS congruence)
Hence, $\angle ACD=\angle ACB={30}^{\circ }$ [C.P.C.T.C.]