In square ABCD, BD and AQ intersect at P with QD = 1 and the area of triangle ABP is equal to AB/2. Find the side of the square.
+9519 Let \(H\in AB\) such that \(PH \perp AB\) and let \(T\in CD\) such that \(PT \perp CD\)
\(\dfrac{AB}2 = \dfrac{AB\cdot PH}{2}\\ PH = 1\)
Let x be the side length of the square and considering similar triangles,
\(x = \dfrac1{x - 1}\)
I will leave the solving process for you as an exercise.
At the end, we get a beautiful result, which is \(x = \varphi\), where \(\varphi\) is the golden ratio.
