# A Problem in Pentagon with Right Angles

### Solution 1

We'll use complex numbers, using lower case $x\,$ to represent point $X.\,$ From the condition on two angles and the geometric meaning of multiplication by $i,$

$d-c=i(b-c),\\ d-e=-i(a-e).$

Subtracting the second from the first and dividing by $2\,$ we get

$\displaystyle \frac{e-c}{2}=i\frac{a+b}{2}-i\frac{c+e}{2}$

which is the same as

$\displaystyle h=\frac{c+e}{2}+i\frac{c-e}{2}.$

This exactly means that $\Delta CEH\,$ is right isosceles, $EH=CH\,$ and $\angle CHE=90^{\circ}.$

### Solution 2

Consider a sequence of transformations: first rotate $90^{\circ}\,$ around $E\,$ (taking $A\,$ to $D),\,$ then rotate $90^{\circ}\,$ about $C\,$ (taking $D\,$ to $B),\,$ and finally rotate $180^{\circ}\,$ about $H\,$ taking $B\,$ to $A.\,$

The resulting transformation, which is either the identity or a translation, must be the identity since it takes $A\,$ to $A.$

Now follow point $E:\,$ $E\,$ is fixed by the first rotation, then goes to $E',\,$ which then reflects across $H\,$ back to $E.$

Then $\Delta ECE'\,$ is an isosceles right triangle, and $CH\,$ is the median to the hypotenuse. Therefore, $CH\perp EH\,$ and $CH=EH.$

Let $M\,$ be the midpoint of $AD,\,$ $N\,$ that of $BD.$

Then $\Delta HME=\Delta CNH.$

### Acknowledgment

The problem was kindly posted at the CutTheKnotMath facebook page by Marian Dinca, with a suggestion to employ complex numbers for a short solution (Solution 1). Solution 2 is by Ted Courant. The problem appeared as Problem 1049 at the Οι Ρομαντικοι της Γεωμετριας (Romantics of Geometry) facebook group.

### Remark

Thanos Kalogerakis came up with a generalization that sheds light on the essence of the problem:

In a pentagon $ABCDE,\,$ $\angle AED+\angle BCD=180^{\circ}\,$ and also $BC=CD\,$ and $AE=DE.$

Prove that

1. The point $K\,$ of intersection of the two circles $(ADE)\,$ and $(BCD)\,$ is on $AB;$

2. Circle $(CEK)\,$ passes through the midpoint $H\,$ of $AB.$

We shall discuss this problem ">later on.