# Prove that $|PC|^2 + |PD|^2 = |AB|^2$ if

We have an angle of 90° so that there are 2 points A, B on each side of the angle, O is the vertex and |OA| = |OB|. On the arc AB with it's center being in O, we pick an arbitrary point P and draw a line through P so that the line is parallel to the line AB. This line touches OA in C, and OB in D. Now prove that: $|PC|^2 + |PD|^2 = |AB|^2$

Since the point P is arbitrary and there is no mention of the line CD being a tangent, I concluded that it had to look like this: http://img843.imageshack.us/img843/9205/2532013114412.png

So I started to write out some relations: $|OA|^2 + |OB|^2 = |AB|^2$ and since $|OA| = |OB|$ we get $2|OA|^2 = |AB|^2$ and since the AB and CD are parallel we can use some proportionality properties (Thales): $\frac{|OA|}{|OC|}=\frac{|OB|}{|OD|}$ and since |OA| = |OB| we conclude: |OC| = |OD| and then we have also: $|OC|^2 + |OD|^2 = |CD|^2$ which can be written as $2|OC|^2 = (|CP| + |PD|)^2 \implies 2|OC|^2 = |CP|^2 + 2|CP||PD| + |PD|^2 \implies 2|OC|^2 - 2|CP||PD| = |CP|^2 + |PD|^2$

so I would have to prove that $2|OA|^2 = 2|OC|^2 - 2|CP||PD|$

but I don't know what to do next, I added some points and tried to find some relations but no luck. It would be great if someone could give me a hint.

NOTE: I put the analytical tag because the course description includes analytical geometry but as for now, we use only elementary geometry i.e. Thales, Pythagoras, theorems of the similarity/coherence of 2 triangles, the surface of triangles and quadrilaterals, theorems for circles (the angle over an arc) and the properties of tangential quadrilaterals and cyclic quadrilaterals but no trigonometry properties (we don't use any trigonometry as for now).

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Since $AB \parallel CD$, triangle $OAB$ is similar to triangle $OCD$, so $|OC|=|OD|$. Let $M$ be the foot of the perpendicular from $O$ onto $CD$. Triangles $MCO, OCD, MOD$ are similar, so $|CM|=|DM|=|OM|$. \begin{align} |PC|^2+|PD|^2& =(|MC|-|PM|)^2+(|MD|+|PM|)^2 \\ & =(|OM|-|PM|)^2+(|OM|+|PM|)^2 \\ & =2|OM|^2+2|PM|^2 \\ & =2|OP|^2 \\ & =|AB|^2 \end{align}

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Thank you very much! :) –  Shirohige Mar 25 '13 at 16:33

Using coordinate geometry gives the result without too much difficulty. Let the circle have radius $r$. Then we can write $P$ as $(r\cos\theta,r\sin\theta)$, and we can write this equation for the line $CD$: $x+y=r\cos\theta+r\sin\theta$. (Note that the line through $A,B$ has slope -1.) It is straightforward from there to work out the coordinates of $C$ and $D$, calculate the lengths required and so prove the result.

Of course you may need a synthetic Euclidean proof for which this isn't much use, but with 'analytic geometry' in the tags, the coordinate approach may be sufficient.

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I am not allowed to use this, I put the analytical tag because the course description includes analytical geometry but as for now, we use only elementary geometry i.e. Thales, Pythagoras, theorems of the similarity/coherence of 2 triangles, the surface of triangles and quadrilaterals, theorems for circles (the angle over an arc) and the properties of tangential quadrilaterals and cyclic quadrilaterals but no trigonometry properties (we don't use any trigonometry as for now). –  Shirohige Mar 25 '13 at 11:39
And yes, I need a hint for the Euclidean proof, I am sorry that I forgot to mention it but still thank you for your answer! –  Shirohige Mar 25 '13 at 11:52

Let angle COP be $\theta$. Since triangles OCD and OAB are similar, the angle OCD is $\pi/4$. Applying the sine rule to triangle OCP we get $|CP|/\sin\theta=|OP|/sin(\pi/4)$ which gives $|CP|=|OP|\sqrt2 \sin\theta$. Similarly for the other side we get $|PD|=|OP|\sqrt2\sin(\pi/2-\theta)=|OP|\sqrt2\cos\theta$. Using these expressions for |CP| and |PD| we get $|CP|^2+|PD|^2=2|OP|^2(sin^2\theta+cos^2\theta)=2|OA|^2=|AB|^2$ as required.

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Thank you for your answer but I am not allowed to use trigonometry, I forgot to mention it so I edited the original post. I am sorry! –  Shirohige Mar 25 '13 at 11:50