The Parallel Proof
**Nicole Mosteller**
**EMAT 6690**

**This proof requires the application
of parallel line properties and isosceles triangle properties.**

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

First, construct a line through point
M that is parallel to segment ZY.
__Figure 2:__ PM is parallel to ZY.

By extending a common parallel property,
we know that if a segment bisects
one side of a triangle (side XY) and
is parallel to another side of the triangle (side ZY),
then it must also bisect the third
side of the triangle (side XZ).
And so, XP is congruent to ZP.
__Figure 3:__ MP bisects XZ.

Since segment MP is parallel to segment
ZY (by construction) and segment ZY is perpendicular to segment
XZ,
then segment MP must also be perpendicular
to segment XZ.
By definition, MP is an altitude of
triangle XMZ
__Figure 4:__ Triangle XMP is isosceles.
Because MP is both a bisector and an
altitude of triangle XMP, we know that triangle XMP is isosceles
with side XM congruent to side ZM.

Segment ZM is congruent to segment
XM (legs of isosceles triangle), and
segment XM is congruent to segment
YM (definition of median).
By transitive property, ZM = XM = YM.
__Figure 5:__ ZM = XM = YM.
So, the midpoint M is eqidistant from
all of the vertices of the right triangle.

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