Question
A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
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Answer
Let $\triangle\text{ABC}$ be an isosceles right triangle, right-angled at B.$\Rightarrow\text{AB = BC}$
Let PBSR be a square inscribed in $\triangle\text{ABC}$ with common $\angle\text{B}.$
$\Rightarrow\text{PB = BS = SR = RP}$
Now, $\text{AB} - \text{PB = BC} -\text{BS}$
$\Rightarrow\text{AP = CS ...(i)}$
In $\triangle\text{APR}$ and $\triangle\text{CSR}$
$\text{AP = CS}$ [from (i)]
$\angle\text{APR}=\angle\text{CSR}$ (Each 90°)
$\text{PR = SR}$ (sides of a square)
$\therefore\triangle\text{APR}\cong\triangle\text{CSR}$ (by SAS congruence criterion)
$\Rightarrow\text{AR = CR}$ [C.P.C.T.]
Thus, point R bisects the hypotenuse AC.
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