Question
In a quadrilateral ABCD, $\angle\text{B}=90^\circ.$ If $AD^2 = AB^2 + BC^2 + CD^2$, then prove that $\angle\text{ACD}=90^\circ.$
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Answer
Given: In quadrilateral ABCD, $\angle\text{B}=90^\circ$ and $AD^2 = AB^2 + BC^2 + CD^2$
To prove: $\angle\text{ACD}=90^\circ$
Proof: In $\triangle\text{ABC},\ \angle\text{B}=90^\circ$
$AC^2 = AB^2 + BC^2 .....(i)$
$But AD^2 = AB^2 + BC^2 + CD^2$
$\Rightarrow AB^2 + BC^2 = AD^2 - CD^2 .....(ii)$
Substituting the value of $AB^2 + BC^2$ in (i)
$AC^2 = AD^2 - CD^2$
$\Rightarrow AD^2 = AC^2 + CD^2$
$\therefore\triangle\text{ACD}$ is a right triangle
Hence $\angle\text{ACD}=90^\circ$
To prove: $\angle\text{ACD}=90^\circ$
Proof: In $\triangle\text{ABC},\ \angle\text{B}=90^\circ$
$AC^2 = AB^2 + BC^2 .....(i)$
$But AD^2 = AB^2 + BC^2 + CD^2$
$\Rightarrow AB^2 + BC^2 = AD^2 - CD^2 .....(ii)$
Substituting the value of $AB^2 + BC^2$ in (i)
$AC^2 = AD^2 - CD^2$
$\Rightarrow AD^2 = AC^2 + CD^2$
$\therefore\triangle\text{ACD}$ is a right triangle
Hence $\angle\text{ACD}=90^\circ$
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