If we write QED it means it is GEOMETRY ??

 QED = GEOMETRY

More and more people are hesitant to use the expression QED at the end of a demonstration.

          Regarding problem E:17017 from GMB 10/2024.

In translation:

               "E:17017. Let $ABC$ be any triangle with $AB=3 \cdot BC$. Consider the 

 points $D\in AC$ such that $AD= 3 \cdot DC$, and $E$ the midpoint of the side $AB$.

 Show that $BD \perp DE$."

Solution CiP

          Due to the dimensions given in the problem it is convenient to choose $DC=p$ and $AB=6\cdot m$, hence $AD=3 \cdot p,\;BE=AE=3\cdot m,\;BC=2 \cdot m$. 

          It is obvious that $\frac{AE}{AC'}=\frac{3m}{4m}=\frac{3}{4}=\frac{3p}{4p}=\frac{AD}{AC}$ so with the Reciprocal of Thales' Theorem

$$DE \parallel CC' \tag{1}$$

          It is also obvious that $\frac{BC}{BA}=\frac{2m}{6m}=\frac{1}{3}=\frac{p}{3p}=\frac{CD}{AD}$ so with the Converse of Angle Bisector Theorem $BD$ is the bisector of angle $\measuredangle CBA$.

Then, in the obvious isosceles triangle $BCC'$ the bisector of angle at vertex $B$ is also the altitude, so $BD \perp CC'$. From (1) it then follows $BD \perp DE$.

QED

$\blacksquare$

            Remark CiP

          A more computational solution uses the Law of Cosines.

     In $\Delta ABC\;: \quad cos\; C=\frac{CB^2+CA^2-AB^2}{2\cdot CB \cdot CA}=\frac{4m^2+16p^2-36m^2}{2\cdot 2m\cdot 4p}=\frac{p^2-2m^2}{mp}$

and in $\Delta BCD\;:\quad BD^2=CB^2+CD^2-2\cdot CB \cdot CD \cdot cos\;C=$

$=4m^2+p^2-2 \cdot 2m \cdot p \cdot \frac{p^2-2m^2}{mp}=4m^2+p^2-4(p^2-2m^2)$ so

$$BD^2=12m^2-3p^2 \tag {2}$$

Again in $\Delta ABC\;:\quad cos\;A=\frac{AB^2+AC^2-BC^2}{2\cdot AB \cdot AC}=\frac{36m^2+16p^2-4m^2}{2\cdot 6m \cdot 4p}=\frac{p^2+2m^2}{3mp}$

and in $\Delta ADE\;: \quad DE^2=AD^2+AE^2-2\cdot AD \cdot  AE \cdot cos\;A=$

$=9p^2+9m^2-2\cdot 3p \cdot 3m \cdot \frac{p^2+2m^2}{3mp}=9p^2+9m^2-6(p^2+2m^2)$ so

$$DE^2=3p^2-3m^2 \tag{3}$$

From (2) and (3) it result

$$BD^2+DE^2=(12m^2-3p^2)+(3p^2-3m^2)=9m^2=BE^2$$

hence by Converse of Pythagorean Theorem the angle between sides $BD$ and $DE$ is a right angle.

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