In a square PQRS, diagonals PR and QS intersects at O. The angle bise…

Let side PQ = a

So diagonal PR = \(\sqrt {2}\)a

⇒ OP = OQ = OR = OS = \(\frac{PR}{2}\) = \(\frac{a}{\sqrt {2}}\)  .........(1)

Now, in ΔPOQ and from angle bisector theorem

PQ : PO = QT : OT

a = \(\frac{a}{\sqrt {2}}\) = QT : OT

\(\frac{\sqrt {2}}{1}\) = \(\frac{QT}{OT}\)

Let QT = \(\sqrt {2}\)b and OT = b  ..........(2)

QT + OT = OQ = \(\frac{a}{\sqrt {2}}\)      [from (1)]

\(\sqrt {2}\) b + b = \(\frac{a}{\sqrt {2}}\)

⇒ b = \(\frac{a}{\sqrt {2}\left(\sqrt {2} +1\right)}\)

OT = \(\frac{a}{\sqrt {2}\left(\sqrt {2} +1\right)}\)  ..........(3)

Now, in ΔPQR and from angle bisector theorem

PQ : PR = QU : RU

a : \(\sqrt {2}\)a = QU : UR

1 : \(\sqrt {2}\) = QU : UR

at QU = c, UR = \(\sqrt {2}\)c  .........(4)

Let QU + RU = QR = d

c + \(\sqrt {2}\)c = d

c = \(\frac{d}{\sqrt {2}+1}\) ⇒ \(\sqrt {2}\)c = \(\frac{\sqrt {2}d}{\sqrt {2}}\)  ..........(5)

Now from (3) and (5)

OT : RU = \(\frac{a}{\sqrt {2}\left(\sqrt {2} +1\right)}\) : \(\frac{\sqrt {2}a}{\sqrt {2}+1}\)

= \(\frac{1}{\sqrt {2}}\) : \(\frac{\sqrt {2}}{1}\)

= 1 : 2