I'm writing a paper at the moment, and I've stumbled across a fact that I need. I can prove it analytically, using inner products, but I especially want a simple Euclidean geometry proof.
The fact is as follows: suppose we have two lines that intersect uniquely at a point $B$, and fix another point $D$ on the plane. Let $G$ and $F$ be the respective points of orthogonal projection, from $B$, onto the two lines. Further, let $K$ and $H$ be points on $DG$ and $DF$ respectively, such that $\vec{DK}$ is the unit vector in the direction of $\vec{DG}$ and $\vec{DH}$ is the unit vector in the direction of $\vec{DF}$.
Then, as I've discovered, $$\left|\vec{DB}\right| = \frac{2 \cdot \left|\vec{GF}\right|}{\left|\vec{DK} + \vec{DH}\right| \cdot \left|\vec{DK} - \vec{DH}\right|}.$$
I've got a messy, inelegant and uninformative proof using inner products, so I was hoping to find a proof that just uses simple Euclidean geometry.
Anyone have some ideas?
Let $\angle GDF=\theta$. Then $\displaystyle \left|\overrightarrow{DK}+\overrightarrow{DH}\right|=2\cos\frac{\theta}{2}$ and $\displaystyle \left|\overrightarrow{DK}-\overrightarrow{DH}\right|=2\sin\frac{\theta}{2}$
So, $$\frac{2}{\left|\overrightarrow{DK}+\overrightarrow{DH}\right|\left|\overrightarrow{DK}-\overrightarrow{DH}\right|}=\frac{2}{2\sin\theta}=\frac{1}{\sin\theta}$$
Let $O$ be the midpoint of $BD$. It is the centre of the circle $BDGF$. Let $M$ be the midpoint of $GF$. Then $\displaystyle \angle FOM=\frac{\angle GOF}{2}=\frac{2\angle GDF}{2}=\theta$
So, $\displaystyle \sin\theta=\frac{\left|\overrightarrow{MF}\right|}{\left|\overrightarrow{OF}\right|}=\frac{\frac{1}{2}\left|\overrightarrow{GF}\right|}{\frac{1}{2}\left|\overrightarrow{DB}\right|}=\frac{\left|\overrightarrow{GF}\right|}{\left|\overrightarrow{DB}\right|}$
Therefore, $$\left|\overrightarrow{DB}\right|=\frac{2\left|\overrightarrow{GF}\right|}{\left|\overrightarrow{DK}+\overrightarrow{DH}\right|\left|\overrightarrow{DK}-\overrightarrow{DH}\right|}$$