The Similarity Proof
**Nicole Mosteller**
**EMAT 6690**

**This proof requires applications
of reflection of a triangle over a line and similar triangles.**

Given a right triangle XYZ with the
median of the hypotenuse labeled M.
__Figure 1:__ Given.

Using property of reflection, reflect
triangle XYZ across segment YZ.
Figure 2: Reflection of triangle XYZ.

The reflection of M is the point M'.
When these points are connected, we have the
the additional segment MM' in our figure.
__Figure 3:__ Segment MM'.

Now, let's observe the various triangles
created by the reflection.
Triangle MYM' is similar to triangle
XYX' (by SAS).
Since M is the midpoint of segment
XY,
these triangles are similar by a scale
factor of 1:2
__Figure 4:__ MYM' is similar to XYX'.

Triangle XMZ is similar to triangle
XYX' (by SAS as well).
The reflection of X to X' also makes
these triangles similar by a
scale factor of 1:2.
__Figure 5:__ XMZ is similar to XYX'.

Since triangle MYM' and triangle XMZ
are similar to the same triangle and
because the similarities are with the
same scale factor 1:2,
then triangle MYM' must be congruent
to triangle XMZ.
__Figure 6:__ MYM' is congruent to XMZ.
This triangle congruence leads to the
following list of congruences:
ZM = M'Y (CPCTC), M'Y = MY (reflected),
MY = MX (def. of median).

By the transitive property, ZM = MY
= MX.
__Figure 7:__ ZM = MY = MX.
So, the midpoint M is eqidistant from
all of the vertices of the right triangle.

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