Please help me with this problem.
Hi
Consider triangle BPE. PR is one height of this triangle. The area of BPE is $\displaystyle A(BPE) = \frac12\:PR \times BE$
The same for triangle BPC : $\displaystyle A(BPC) = \frac12\:PQ\times BC$
The area of triangle BCE is the sum of the areas above : $\displaystyle A(BCE) = A(BPE) + A(BPC) = \frac12\:PR\times BE + \frac12\:PQ\times BC$
Since BE = BC, $\displaystyle A(BCE) = \frac12\PQ + PR)\times BC$
On the other hand the area of triangle BCE can be calculated using the projection of E on [BC], say H
$\displaystyle A(BCE) = \frac12\:EH\times BC$
Since ABCDE is a square, angle CBE is 45 degrees. Therefore $\displaystyle EH = BE\times \sin(45) = BC \:\frac{\sqrt{2}}{2}$
$\displaystyle A(BCE) = \frac12\:\frac{\sqrt{2}}{2}\times BC^2$
Hence $\displaystyle A(BCE) = \frac12\PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2$
And finally $\displaystyle PQ + PR = \frac{\sqrt{2}}{2}\times BC$