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