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I've stumbled upon the following problem that made me curious for quite a while. Below is one of many possible diagrams of the problem in question.

The given diagram

Given that $ABCD$ is a parallelogram with acute angles $\angle A$ and $\angle C$, $E$ is the midpoint of $BC$, and that lines through $B$ perpendicular to $AB$ and through $E$ perpendicular to $DE$ intersect at point $K$, I have to prove that $KA = KD$.

One approach I know to work is the following:

  • Prolong $AB$ and $DE$ to intersect at some point $P$. These two lines are guaranteed to intersect for the following reason: $AB\parallel DC$, $BC\cap DC = C$, $E\in BC$, $E\not\equiv C$, therefore $E\notin\overleftrightarrow{DC}$. Thus, $\overleftrightarrow{DE} \cap \overleftrightarrow{DC} = D$. With $AB \parallel DC$, we conclude that $DE \not\parallel AB$, as through a given point not on a given line passes at most one line parallel to the given line.
  • Then triangle $\triangle APD$ has $BE$ as a midline parallel to $AD$. That is because $ABCD$ is a parallelogram with $BC\parallel AD$, so $BE\parallel AD$, and since $E$ is a midpoint of $BC$, and $BC = AD$, then $AD = 2\cdot BE$.
  • Therefore, $KB$ and $KE$ are perpendicular bisectors of triangle $\triangle APD$, and from there we conclude $K$ is a circumcenter of triangle $\triangle APD$.
  • It follows that $KA = KD$ as the radii of the circumcircle of triangle $\triangle APD$, which was to be proven.

I am exploring a more elegant way to achieve the same result.

  • Let $M$ be the midpoint of $AD$. Construct segment $ME$. Segment ME
  • As $BE = AM$ by construction of points $E$ and $M$, we have $BE = AM$ and $BE\parallel AM$, therefore $ABEM$ is a parallelogram, from where $AB\parallel ME$.
  • $BK\perp AB$, $ME\parallel AB$, therefore $BK\perp ME$ as well. BK perpendicular to ME
  • Construct segment $KM$ by connecting points $K$ and $M$. Construct a segment KM

I now observe that $MKED$ is a cyclic quadrilateral. MKED is cyclic

However, this is where I am stuck. I don't want to resort to the first proof I gave to prove $KM\perp AD$, so I wanted to prove that $M$, $K$, $E$, and $D$ are concyclic, and from there I'd conclude that $KM\perp AD$, and get the necessary result from the properties of perpendicular bisector.

How do I prove that $M$, $K$, $E$, and $D$ are concyclic?

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3 Answers 3

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Let $\Gamma$ be the circle through $K,E,D$. Define points $X = BK \cap CD$ and $Y = BK \cap EM$.

From equal heights of parallelograms $BEMA$ and $ECDM$, we have $BY=YX$. Hence, by $SAS$ congruence we establish $\triangle BYE \equiv \triangle XYE$, and hence $BE = EX=MD$.

Thus, as $EX=MD$ and also $EM \parallel XD$ we have $EMDX$ as an isosceles trapezium and hence cyclic.

Furthermore, from alternate angles, $90^\circ = \angle ABY = \angle YXC \Longrightarrow \angle KXD = 90^\circ$.

Hence, as $\angle KED = \angle KXD = 90^\circ$, by Bowtie Theorem, $KEXD$ is cyclic.

Finally, we have that $EMDX$ and $KEXD$ share the same circle, so $MKEXD \Rightarrow MKED$ cyclic.

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  • $\begingroup$ Wonderful answer! I haven't initially noticed the obvious equality of $ABEM$ and $MECD$. The rest of the reasoning is rather mechanical as soon as $BY = YX$ is noticed. Angle $\angle YXB$ is $0^\circ$ though, as $B, X, Y$ are collinear. $\endgroup$ Commented 19 hours ago
  • $\begingroup$ Whoops! Fixed the typo :) $\endgroup$ Commented 10 hours ago
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Since $ABCD$ is a parallelogram and $E, M$ are the midpoints of $BC, AD$ respectively, $EM \parallel AB,\; BE \parallel AD,\; ED \parallel BM$.

Given $BK \perp AB \Rightarrow BK \perp EM$.

Similarly, $EK \perp ED$ and $ED \parallel BM \Rightarrow EK \perp BM$.

Thus $BK, EK$ are altitudes of $\triangle BEM$, hence $K$ is its orthocenter.

Therefore $MK \perp BE$.

Since $BE \parallel AD \Rightarrow MK \perp AD \Rightarrow MK \perp MD$, so $\angle KMD = 90^\circ$.

Also $EK \perp ED \Rightarrow \angle KED = 90^\circ$.

Hence $\angle KED = \angle KMD = 90^\circ \Rightarrow K, E, D, M$ are concyclic.

enter image description here

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HINT.-In your five figures angle $\angle{KED}$ is right so in your last figure, points $K,E,D$ (whose coordinates are known) determine a circle whose diameter is $\overline{KD}$. One way to finish is to verify that point $M$ belongs to this circle which is easy to do.

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  • $\begingroup$ $\angle AED$ is not a right angle because $A$, $K$, and $E$ are non-collinear. $\endgroup$ Commented 21 hours ago
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    $\begingroup$ @TheProver: Your are correct, the angle is $\angle{KED}$ (it was a type). $\endgroup$ Commented 21 hours ago

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