Median of a right isosceles triangle
To show that the median of a right isosceles triangle is half the hypotenuse, start with a right triangle and then draw the median (shown in red)
First, show that triangle ABD and triangle ADC are congruent. If we can show this, then we can conclude that angles BAD and angle DAC are congruent and each is equal to 45 degrees
This will also mean that angle BDA = 90 degrees. Therefore, we can use the Pythagorean Theorem for triangle ABC and triangle and triangle ABD
Let us show that triangle ABD and triangle ADC are congruent by SSS (three equal sides)
We already know that segment AB = segment AC since triangle ABC is isosceles
Since segment AD is the median of segment BC, segment BD = segment DC
Finally, segment AD is a common side, so it is equal for both triangles
We have found 3 equal sides, so triangle ABD and triangle ADC are congruent and we can do the aforementioned
Now, let's use the Pythagorean Theorem for triangle ABC and triangle and triangle ABD. The focus now will be on lengths of sides
For triangle ABC,
x
^{2} = y
^{2} + y
^{2}
x
^{2} = 2y
^{2}
y
^{2} = (x
^{2}) / 2
For triangle ABD, y
^{2} = z
^{2} + (x/2)
^{2}
Substitute (x
^{2}) / 2 for y
^{2}
We get:
(x
^{2}) / 2 = z
^{2} + (x/2)
^{2}
(x
^{2}) / 2 = z
^{2} + (x
^{2}) / 4
(x
^{2}) / 2  (x
^{2}) / 4 = z
^{2}
(2x
^{2}) / 4  (x
^{2}) / 4 = z
^{2}
(x
^{2}) / 4 = z
^{2}
x/2 = z (Done!)
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Sep 11, 17 05:06 PM
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