1. ## Circles

Let $\displaystyle \bold{r}$ be a point on a circle with diameter from $\displaystyle - \bold{a}$ to $\displaystyle \bold{a}$ (so that $\displaystyle |\bold{r}| = |\bold{a}|$). Draw the chord through $\displaystyle \bold{r}$ perpendicular to $\displaystyle \bold{a}$ (the length and direction of this chord is $\displaystyle 2 \bold{r}_{a \perp}$). Let $\displaystyle \bold{t} = \bold{r}_{a}$ be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from $\displaystyle -\bold{a}$ to $\displaystyle \bold{t}$ times the distance from $\displaystyle \bold{t}$ to $\displaystyle \bold{a}$: $\displaystyle |\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t})$.

2. So I got $\displaystyle \hat{\bold{a}} = \frac{\bold{a}}{|\bold{a}|}$

and $\displaystyle \bold{t} = \bold{r} \cdot \hat{\bold{a}}$

and $\displaystyle \bold{r}_{a \perp} = \bold{t} - \bold{r}$.

Then $\displaystyle \bold{a} + \bold{r}$ gives one hypotenuse. And so $\displaystyle |\bold{a} + \bold{r}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}+\bold{t}|^{2}$.

So $\displaystyle |\bold{r} - \bold{a}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}-\bold{t}|^{2}$.

3. Originally Posted by heathrowjohnny
Let $\displaystyle \bold{r}$ be a point on a circle with diameter from $\displaystyle - \bold{a}$ to $\displaystyle \bold{a}$ (so that $\displaystyle |\bold{r}| = |\bold{a}|$). Draw the chord through $\displaystyle \bold{r}$ perpendicular to $\displaystyle \bold{a}$ (the length and direction of this chord is $\displaystyle 2 \bold{r}_{a \perp}$). Let $\displaystyle \bold{t} = \bold{r}_{a}$ be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from $\displaystyle -\bold{a}$ to $\displaystyle \bold{t}$ times the distance from $\displaystyle \bold{t}$ to $\displaystyle \bold{a}$: $\displaystyle |\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t})$.
Hello,

you are dealing with a right triangle (see attachment)

If the legs of a right triangle are a and b and the hypotenuse is c then you know:

$\displaystyle a^2+b^2 = c^2$ ....... and $\displaystyle c = p+q$. That means $\displaystyle c^2 = p^2+2pq+q^2$

Therefore:

$\displaystyle a^2+b^2 = p^2+2pq+q^2$

According to the drawing you have:

$\displaystyle a^2 = h^2 + p^2$ ....... and $\displaystyle b^2 = h^2 +q^2$

So finally you get:

$\displaystyle h^2 + p^2 + h^2 +q^2 = p^2+2pq+q^2~\iff~ 2h^2 = 2pq~\iff~\boxed{h^2 = pq}$