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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:
- ^@\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 }].^@
