3
$\begingroup$

We're given an angle $\angle BAC$. We want to construct $D \in \overrightarrow{AB}$ and $E \in \overrightarrow{AC}$ such that $CE=DE=DB$. This is Euclidea's problem 15.4 and I have the solution but I don't understand it, here is a solution:

enter image description here

we construct circle $\gamma = \odot (B,\overline{AC})$ and mark point $X = \gamma \cap \overline{AB}$

we then draw line $r$, such that $X \in r \parallel AC$

we draw circle $\omega = \odot (C,\overline{CB})$ and mark point $Y = \omega \cap r$

point $D$ is in the perpendicular bissector of $YB$ and on line $AB$.

Why is that?

People who want solutions for Euclidea can check this channel

$\endgroup$
1
  • 1
    $\begingroup$ The key point is to show that CED is part of a rhombus. $\endgroup$
    – Mick
    Apr 1, 2023 at 5:31

2 Answers 2

2
+50
$\begingroup$

I am trying to answer the question "Why is that?" in the following manner. First of all, assuming we have the points $D,E$ on the sides $AB,AC$ with $BD=DE=EC$ we will construct $Y,X$, and show their properties match those from the construction. So if there is a solution, it must follow the prescribed receipt. Secondly, we perform the prescribed receipt (from the question and from the first point) till the end, also introducing $E$ in the obvious unique manner, and show that $BD=DE=EC$. So we have two proposition, let us state them explicitly in detail.

The points $A,B,C$ are given. Furthermore, points $D$ on the segment $AB$, $E$ on $AC$ are given, so that $BD=DE=EC$. Consider $Y$, the reflection of $B$ w.r.t. $CD$. Draw the parallel $r$ through $Y$ to $AC$, it intersects $AB$ in a point $X$.

Then: $XD=AE$ and $XB=AC$.

Proof: Denote by $a$ and by $2e$ the (measures of) angles $a=\widehat{BAC}$, and $2e=\widehat{CED}$.

Construction euclidea problem MSE 4670220

Then the angles in $C,D$ in the isosceles triangle $\Delta CED$ are the complements of $e$, $\hat C=\hat D=\bar e:=90^\circ-e$. Also, $\widehat{EDA}=\widehat{CED}-\widehat{BAC}=2e-a$. So $\widehat{CDA}= \bar e+(2e-a)=90^\circ+e-a$. This is also half of the angle in $D$ in $\Delta DBY$, so the equal angles in $B,Y$ in this last triangle are each $a-e$. The angle $a=\widehat{AXY}$ is finally the sum of $\widehat{XYB}$ and $\widehat{XBY}$, so $$ \widehat{XYD}= a- 2(a-e)=2e-a=\widehat{EDA}\ . $$ To conclude now that $EA=XD$ the simplest way is to use the sine theorem for $\Delta EAD$, and $\Delta DXY$: $$ \frac{EA}{ED}= \frac{\sin (2e-a)}{\sin a} = \frac{\sin (2e-a)}{\sin (180^\circ-a)} = \frac{XD}{DY}\ . $$ Now we use $ED=DB=DY$ to get $EA=XD$, and finally $AC=AE+EC=XD+DB=XB$.

$\square$

The points $X,Y$ correspond to those from the question.


Conversely, we show that the construction from the question is working. (At each point, only properties of the construction may be used as arguments.) Here is explicitly the complete statement.

Consider the points $A,B,C$ in the plane. We assume (to fit the figure) $AB\ge AC$. Construct first the point $A$ on $AB$ such that $AC=BX$. Construct the parallel $r$ through $X$ to $AC$, and let $Y$ be the point on it with $CY=CB$. (There are in general two such points. In order to match the figure from the question, we make an assumption.) We assume that $Y$ is in the half-plane w.r.t. $AB$ that does not contain $C$. The perpendicular bisector of the segment $BY$ intersects $AB$ in a point $D$. In particular $DB=DY$. The perpendicular bisector of $CD$ intersects $AC$ in a point $E$.

Then: $BD=DE=EC$.

Proof: The picture is as follows:

Construction euclidea problem MSE 4670220

Again, we denote by $a$ the angle $\widehat{BAC}$, by $2e$ the angle $\widehat{CED}$. Then the angles in $C,D$ in $\Delta CED$ are $\bar e$, $2e$ is exterior to $\Delta AED$ in $E$, so $\widehat{EDA}=2e-a$, this leads to the value of the half of $\hat D$ in the isosceles triangle $\Delta DBY$, so the angles in $B,Y$ are each $a-e$. Finally, since $a=\widehat{BAC}=\widehat{AXY}$, we compute $\widehat{XYD}=2e-a$. The sine theorem in $\Delta AED$ and $\Delta XDY$ gives now the needed transition from proportions on $AB$ to proportions on $AC$: $$ \frac {XD}{DB} = \frac {XD}{DY} = \frac {\sin (2e-a)}{\sin(180^\circ -a)} = \frac {\sin (2e-a)}{\sin a} = \frac {AE}{ED} = \frac {AE}{EC}\ . $$ From here, adding one on each side, $$ \frac {XB}{DB} = \frac {XD}{DB} +1= \frac {AE}{EC} +1= \frac {AC}{EC}\ . $$ But by construction, $XB=AC$, so $DB=EC$.

$\square$

$\endgroup$
1
$\begingroup$
  1. Construct the ||gm CABZ so that circle $γ(B, AC=BZ)$ can be drawn.

enter image description here

  1. Produce EC to V such that EC = CV.

  2. Produce ED to cut the red dotted circle $\omega$ at R. Note that ED = DR = radius of $\omega$.

  3. Form the $\triangle EVR$ and let VR cut $\omega$ at U.

  4. By midpoint theorem, CD || VR.

  5. Let EU cut CD at S.

  6. By angle in semi-circle, $\angle ESC = \angle SUR = 90^0$.

  7. Since DS is a line from center (D) and is perpendicular to the chord EU, it bisects EU.

  8. With the fact that DE = DU, CEDU is not just a kite but a rhombus.

$\endgroup$
2
  • $\begingroup$ I still need to think how that relates to the original question without assuming the result we want to check ($CE=ED=DB$) $\endgroup$ Apr 1, 2023 at 11:03
  • 1
    $\begingroup$ @hellofriends ED = DB because they are the radii of the same circle. It remains to be shown is whether CE = ED. $\endgroup$
    – Mick
    Apr 1, 2023 at 13:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .