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.

Answer 1

Step-by-step explanation:

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.

Answer 2

Step-by-step explanation:

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=90

o

QR=PQ

∴ΔARQ≅ΔQPB

⇒AQ=QB

∴ Q bisects the hypotenuses