4
$\begingroup$

Let ABCD be a square and Z the midpoint of BC. From vertex A, draw a perpendicular to line DZ with foot E. Draw segment EC. Find ∠ZEC.

Context:

I was studying parallelograms and their properties, especially special cases like squares. Working with square ABCD, I took Z as the midpoint of side BC and drew line DZ. From vertex A, I drew a perpendicular to DZ with foot at point E, then drew segment EC. I tried to find this angle synthetically and wondered if there was some symmetry at play. So I extended AE until it met side CD at point F, and by analyzing the quadrilateral EFCZ that formed, I concluded that the angle equals 45°. I'm interested in seeing other approaches - perhaps using geometric transformations (rotations, reflections), coordinate geometry, or another synthetic proof without auxiliary lines. I'm also curious if there's any generalization of this result.

enter image description here

$\endgroup$
2
  • $\begingroup$ I think you can claim that if $O$ is the midpoint of the square, then $\angle OEC=90^\circ$ and $DZ$ bisects it. $\endgroup$ Commented Dec 12, 2025 at 11:52
  • $\begingroup$ @JMP: Exactly! If we draw $OZ$, then quadrilateral $OECZ$ is cyclic, since $\angle ZOC = \angle ZEC = 45^\circ$. Therefore $\angle OEZ = \angle OCZ = 45^\circ$ and $\angle OEC = 90^\circ$. $\endgroup$ Commented Dec 12, 2025 at 12:57

2 Answers 2

Reset to default
1
$\begingroup$

My synthetic proof:

I extend AE until it meets side CD at point F. I observe that the right triangles ADF and DZC are congruent because they have ∠ADF = ∠DCZ = 90°, AD = DC as sides of the square, and ∠DAF = ∠ZDC since both are acute angles with their sides mutually perpendicular (ASA). Therefore DF = ZC = BC/2, which means F is the midpoint of CD.

In right triangle BCD, points F and Z are the midpoints of sides DC and BC respectively, so it follows that FZ ∥ BD and hence ∠ZFC = ∠BDC = 45° ...(1)

We also observe that quadrilateral EFCZ is cyclic, since ∠FEZ = ∠FCZ = 90° by construction. Therefore, from (1), we conclude that ∠ZEC = ∠ZFC = 45°.

$\endgroup$
1
$\begingroup$

Let $O$ be the centre of the square. Then $O$ is the incentre of $\triangle AEZ$, due to the bisectors of $\angle EAZ$ and $\angle EZA$.

Therefore $\angle OEZ=\frac12 \angle AEZ=45^\circ$.

$\triangle OEF\sim\triangle CED$ (note that $|OF|=\frac12|CD|$ and $OF\perp DC$), therefore $\angle OEC=90^\circ$ and so $\angle ZEC=45^\circ$.

Another proof is that $\triangle DEC\sim\triangle DFZ$ (drop the perpendicular from $E$ to $CD$). As $\angle ZFC=45^\circ$, so is $\angle ZEC$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.