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The point P is a fixed point on the diameter AB of a circle. Show that for any chord CD of the circle (that is parallel to AB) the equation PC^2 +PD^2 = PA^2 + PB^2 is true. P can be anywhere on the diameter.

Feb 6, 2020

#1
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Sorry I accidentally reposted my internet did not save the first one or something. Sorry for the inconvenience. Have a good one!

Feb 6, 2020
#2
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The point P is a fixed point on the diameter AB of a circle.
Show that for any chord CD of the circle (that is parallel to AB) the equation
$$PC^2 +PD^2 = PA^2 + PB^2$$ is true.
P can be anywhere on the diameter. Intersecting chords theorem: $$x*x = y*(PB+PA-y)$$

Pythagorean Theorem:

$$\begin{array}{|lrcll|} \hline (1): & x^2+(PA-y)^2&=&PC^2 \\ (2): & x^2+(PB-y)^2&=&PD^2 \\ \hline \end{array}$$

$$(1)+(2):$$

$$\begin{array}{|rcll|} \hline 2x^2+(PA-y)^2+(PB-y)^2 &=& PC^2+PD^2 \\ \boxed{x^2 = y*(PB+PA-y)} \\ 2y*(PB+PA-y)+(PA-y)^2+(PB-y)^2 &=& PC^2+PD^2 \\ 2yPB+2yPA-2y^2+ PA^2-2yPA+y^2+ PB^2 -2yPB + y^2 &=& PC^2+PD^2 \\ \mathbf{PA^2+ PB^2} &=& \mathbf{PC^2+PD^2} \\ \hline \end{array}$$ Feb 6, 2020
edited by heureka  Feb 6, 2020
edited by heureka  Feb 7, 2020