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)




Right isosceles triangle
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

Right isosceles triangle
For triangle ABC,

x2 = y2 + y2

x2 = 2y2

y2 = (x2) / 2

For triangle ABD, y2 = z2 + (x/2)2

Substitute (x2) / 2 for y2

We get:

(x2) / 2 = z2 + (x/2)2

(x2) / 2 = z2 + (x2) / 4

(x2) / 2 - (x2) / 4 = z2

(2x2) / 4 - (x2) / 4 = z2

(x2) / 4 = z2

x/2 = z (Done!)



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