Trig rules OK
Change the squares in this diagram and spot the property that stays the same for the triangles. Explain...
Problem
Draw any two squares which meet at a common vertex $C$ and join the adjacent vertices to make two triangles $CAB$ and $CDE$.
Construct the perpendicular from $C$ to $AB$, (the altitude of the triangle). When you extend this line where does it cut $DE$?
Now bisect the line $AB$ to find the midpoint of this line $M$. Draw the median $MC$ of triangle $ABC$ and extend it to cut $DE$. What do you notice about the lines $MC$ and $DE$?
Will you get the same results about the two triangles formed if you draw squares of different sizes or at different angles to each other? Make a conjecture about the altitude of one of these triangles and prove your conjecture.
Thank you Geoff Faux for suggesting this problem.
Getting Started
The hint is in the title of the problem.
Student Solutions
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| Curt from Reigate College has a neat way of proving this result by rotating one of the triangles about $C$, keeping the other one fixed, and then using similar triangles.
It is illuminating to see different methods and you may also like to try proving this result using the Sine Rule.
Here is Curt's method.
The question specifies the construction of a line perpendicular to $AB$ drawn through $C$, and continued until it intersects $DE$.
It will be proven that the perpendicular to $AB$ through $C$ bisects $DE$.
To prove this we first develop a useful tool. Note that $\angle BCE=\pi/2$, thus if we were to rotate $B$ by $\pi/2$ radians counter-clockwise to $B'$, then ECB' would be a straight line. Also we note that $\angle ACD= \pi/2$. If we were to rotate $A$ by $\pi/2 $ counter clockwise to $A'$, then $A'$ would coincide with $D$ as $|CD |=|CA|=|CA'|$. |
Image
| The diagram shows the results of rotating triangle $ABC$ by $\pi/2$ counter clockwise.
As lengths between points are invariant under such rotations, $|CA| = |CA'|$, $|AB|=|A'B'|$ and $|BC| =|B'C|$ thus $A'B'C$ is the same triangle as $ABC$. Clearly $EDC$ is unaffected by the rotation; none of its vertices were rotated or translated.
Now in order to see how this is useful, we start from the first diagram, and draw in the extended perpendicular from $Y$ on $AB$ passing through $C$ intersecting $DE$ at $X$. We repeat the rotation about $C$ as shown in the next diagram. |
Image
| In addition to the other transformations already discussed, $Y$ goes to $Y'$. $XCY$ is a straight line perpendicular to $AB$, and $Y$ is rotated $\pi/2$ about $C$ to $Y'$. Thus $XCY'$ is a right angle and $A'B'$ is parallel to $XC$.
Again, as the distances between points that are rotated in the same manner are invariant under rotation, it follows that $|B'Y'|=|BY|$ and $|A'Y'| = |AY|$.
It is out intention to show that $|DX| = |EX|$, or $|DE| = 2|DX|$ |
Teachers' Resources
The problem suggests a property shared by the triangles which always holds no matter how the squares are changed. The challenge is to make and prove a conjecture about this property.