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# help

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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.

Jun 11, 2020

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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.

Jun 11, 2020