Question

If $\angle ACD={90}^{\circ }$ and $CF$ is the angular bisector of $\angle ACD$, then $\angle BCF=$ __${}^{\circ }$.

Answer 1

Given, $\angle ACD$ = ${90}^{\circ }$ and $CF$ is the angular bisector of $\angle ACD$, so $\angle DCF={45}^{\circ }$. So, $BCF={135}^{\circ }$.