Trig Rules OK
Change the squares in this diagram and spot the property that stays
the same for the triangles. Explain...
Age
16 to 18
Challenge level
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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.
The hint is in the title of the problem.
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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 $|CD |=|CA|=|CA'|$.
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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.
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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 $|DE| =
2|DX|$
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Clearly $ECX$ is similar to $EBD'$.
$CX$ is parallel to $DB'$ thus $\angle EB'D = \angle ECX$.
Similarly angle $\angle EXC = \angle EDB'$. Both triangles share
$\angle DEB'$. Thus all three angles in the two triangles are
equal, thus the two triangles are similar.
By the properties of similar triangles, $|EB'|/|EC| =
|ED|/|EX|$. But $|EB'| = 2|EC|$. Thus $ED = 2|EX|$, as
required.
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.