Please help me with this problem.
Hi
Consider triangle BPE. PR is one height of this triangle. The area of BPE is
The same for triangle BPC :
The area of triangle BCE is the sum of the areas above :
Since BE = BC, PQ + PR)\times BC" alt="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
Since ABCDE is a square, angle CBE is 45 degrees. Therefore
Hence PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2" alt="A(BCE) = \frac12\PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2" />
And finally