SOLUTION: in triangle PQR, angle P = 90. PQ= PR, PM is an altitude from P, S is any point on QR such that M is between Q and S, S is between M and R. To prove that QS^2+SR^2=2PS^2
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Question 1038515: in triangle PQR, angle P = 90. PQ= PR, PM is an altitude from P, S is any point on QR such that M is between Q and S, S is between M and R. To prove that QS^2+SR^2=2PS^2
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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Since plane figures are congruent under both rotation and translation, we can place an isosceles right triangle with leg measure 
with the right angle vertex at the origin and the other two vertices at )
and )
without loss of generality.
As will be seen, the relationship of Point S to Point M is immaterial to the theorem to be proven. In fact, we will prove that the required relationship holds for any point S on the segment QR. Therefore, while the coordinates of point M given in the figure can be easily proven, this analysis will be omitted as irrelevant.
The vertices of the two congruent angles in triangle PQR are the intercepts of the line containing the segment QR. Hence, using the Two Intercept Form of the equation of a line, we can derive an equation for the line containing segment QR.
Hence, an arbitrary point on the segment QR has the coordinates:
)
Using the distance formula, we can calculate the measure of segment QS squared:
^2\ +\ \left(a\ -\ (a\ -\ x)\right)^2)
^2\ =\ 2x^2)
Then the measure of SR squared
^2\ +\ (a\ -\ x\ -\ 0)^2)
Then add 
.
I'll leave it as an exercise for you to calculate the square of the measure of segment PS, multiply it by 2, and show that the result is the same as the sum of the squares of the two segments above.
John
My calculator said it, I believe it, that settles it
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