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vineri, 3 ianuarie 2025

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.

<end Rem>


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