Maximum angle of triangle inside a rectangle

Question

Let P be a point on the AB side of the ABCD rectangle. How can I prove the angle CPD is maximum when PA equals PB.

Answer 1

Hint:   let $M$ be the midpoint of $AB$ and $\mathbf{\Gamma}$ the circle through points $M,C,D\,$ which is tangent to $AB$ at $M$ by symmetry.

• The inscribed angle $\angle CMD$ equals $\displaystyle\frac{1}{2}\overparen{CD}\,$, as measured along the arc opposite to $M$.

• Any point $P \in AB$ other than $M$ is strictly outside the circle $\mathbf{\Gamma}\,$, so the secant angle $\angle CPD$ equals $\displaystyle\frac{1}{2}\left(\overparen{CD} - \overparen{P'P''}\right) \lt \frac{1}{2}\overparen{CD}$ where $P',P''$ are the intersections of $PC,PD$ with $\mathbf{\Gamma}$.

Answer 2

let $a,b$ the sidelength of the given rectangle and $$x=AP,y=PB$$ then we get by the theorem of cosines $$\cos(\alpha)=\frac{DP^2+PC^2}{2DP\cdot PC}$$ and by the theorem of Pythagorian we have $$\cos(\alpha)=\frac{b^2+x^2+y^2+b^2}{2\sqrt{x^2+b^2}\sqrt{(a-x)^2+b^2}}=\frac{b^2+x^2+(a-x)^2+b^2}{2\sqrt{x^2+b^2}\sqrt{(a-x)^2+b^2}}=f(x)$$