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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

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