We have, PA ⏊ AC, and RC ⏊ AC

Let AB = a and BC = b

In ΔCQB and ΔCPA

<QCB = <PCA (Common)

<QBC = <PAC (Each 90°)

Then, ΔCQB ~ ΔCPA (By AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or, -----------(i)

In ΔAQB and ΔARC

<QAB = <RAC (Common)

<ABQ = <ACR (Each 90°)

Then, ΔAQB ~ ΔARC (By AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or, -----------(ii)

Adding equation i & ii

=

Or, y () =

Or, y () = 1

Or, =