In a quadrilateral ^@ ABCD ^@, ^@\angle B = 90^\circ^@. If ^@AD | Math Question and Answer | Edugain Europe

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

Answer:

Step by Step Explanation:

$\text{Given}$ : A quadrilateral $ABCD$ in which $\angle B = 90^\circ$ and $AD^2 = AB^2 + BC^2 + CD^2$ .
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, $\angle ACD = 90^\circ [ \text{ By converse of pythagoras' theorem }].$

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