In quadrilateral ABCD,mangle B = 90^{circ}. I | vedclass

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(N/A) Given: In quadrilateral $ABCD$,$\angle B = 90^{\circ}$ and $AD^{2} = AB^{2} + BC^{2} + CD^{2}$.To prove: $\angle ACD = 90^{\circ}$.Proof: Join $

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(N/A) Given: In quadrilateral $ABCD$,$\angle B = 90^{\circ}$ and $AD^{2} = AB^{2} + BC^{2} + CD^{2}$.
To prove: $\angle ACD = 90^{\circ}$.
Proof: Join $AC$. In $\triangle ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$.
Substitute this into the given equation: $AD^{2} = (AB^{2} + BC^{2}) + CD^{2}$.
This becomes $AD^{2} = AC^{2} + CD^{2}$.
In $\triangle ACD$,we have $AD^{2} = AC^{2} + CD^{2}$.
By the converse of the Pythagoras theorem,if the square of one side is equal to the sum of the squares of the other two sides,then the angle opposite to the first side is a right angle.
Therefore,$\angle ACD = 90^{\circ}$.

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

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