### In a quadrilateral $ABCD$, $\angle B = 90^\circ$. If $AD^2 = AB^2 + BC^2 + CD^2$, prove that $\angle ACD = 90^\circ$. B C A D

Step by Step Explanation:
1. $\text{Given}$: A quadrilateral $ABCD$ in which $\angle B = 90^\circ$ and $AD^2 = AB^2 + BC^2 + CD^2$.
2. Here, we have to prove that $\angle ACD = 90^\circ$.

Now, join $AC$.

In $\Delta ABC$, $\angle B = 90^\circ$. \begin{aligned} &\therefore \space AC^2 = AB^2 + BC^2 &&\ldots\text{ (i) } [ \text{By Pythagoras' theorem} ] \\ &\text{ Now }, AD^2 = AB^2 + BC^2 + CD^2 \\ &\implies AD^2 = AC^2 + CD^2 &&[\text{ using(i)}] \end{aligned} Thus, in $\Delta ACD$, we have $AD^2 = AC^2 + CD^2$.

Hence, $\angle ACD = 90^\circ [ \text{ By converse of pythagoras' theorem }].$