# Golden Ratio in Square

In 2015 Tran Quang Hung has found once more the Golden Ratio in a combination of a semicircle, a square, and a right isosceles triangle.

### Construction

Given a right isosceles triangle $ABC$ and its circumcircle, inscribed a square $DEFG$ with a side $FG$ along the hypotenuse $AB.$ Let the side $DE$ extended beyond $E$ intersect the circumcircle at $P.$ Then $E$ divides $DP$ in the Golden Ratio.

This is reminiscent of the golden section by Odom�s construction.

### Proof

A simple construction of the inscribed square leads to a simple calculation giving the ratio $\displaystyle\frac{DP}{DE}=\frac{\sqrt{5}+1}{2},$ the golden ratio. We give a synthetic proof below. From the similarity of the isosceles right triangles $DEC$ and $AEF,$ we have

$\displaystyle\frac{DE}{CE}=\frac{AE}{EF}.$

It thus follows that $DE^{2}=DE\cdot EF = AE\cdot EC.$

If the line $DE$ intersects the semicircle again at $Q,$ then $EQ=DP.$ By the intersecting chords theorem,

$AE\cdot CE=EP\cdot EQ=EP\cdot DP.$

Therefore, $DE^{2}=EP\cdot DP,$ meaning that $E$ divides $DP$ in the Golden Ratio.

The diagram below illustrates the construction of a square inscribed into the right isosceles triangle $ABC.$  ### Golden Ratio 