A square is inscribed in an isosceles right angled 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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Solution

Given: $Δ$ ABC is an isosceles right triangle and square CPQR is inscribed in it.
CPQR is a square
$∴ C P = P Q = P R = R C$
$Δ$ ABC is an isosceles triangle
$∴ A C = B C$
$\Rightarrow A R + R C = C P + B P$
$\Rightarrow A R = B P$ …….. $(1)$ $[∵ R C = C P]$
In $Δ$ ARQ and $Δ Q P B$
$A R = B P$
$∠ A R Q = ∠ Q P B = 90^{o}$
$Q R = P Q$
$∴ Δ A R Q ≅ Δ Q P B$
$\Rightarrow A Q = Q B$
$∴$ Q bisects the hypotenuses AB.

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A square is inscribed in an isosceles right angled 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.

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90oQR=PQ∴ΔARQ≅ΔQPB⇒AQ=QB∴ Q bisects the hypotenuses AB.