trianglestrigonometry
In a right angled triangle, the legs adjacent to the right angle are equal to $a$ and $b$. Prove that the length of the bisector (of the right angle) is equal to $$\frac{a\cdot b\cdot \sqrt{2}}{a+b}.$$
While approaching this question, I was very puzzled as to how I would end up with this expression.
Additionally, I couldn't figure out where the $\sqrt{2}$ would come from, other than the sine or cosine of $45$ degrees (from the bisector).
A method where no trigonometry is used.
Consider triangle $ABC$. Let $AD$, the angle bisector, intersect the circumcircle at $L$. Join $LC$. Consider triangle $ABD$ and triangle $ALC$.
Triangle $ABD$ is similar to triangle $ALC$ (by A.A similarity theorem). Therefore, $$\frac{AD}{AC} = \frac{AB}{AL}$$
i.e, $$AD\cdot AL = AC\cdot AB$$ $$= AD(AD+DL) = AC\cdot AB$$ $$=AD\cdot AD + AD\cdot DL = AC \cdot AB \text{ ... (1)}$$
By power of point of point result: $$AD\cdot DL = BD\cdot DC$$ $$BD = BC\cdot \frac{AB}{AB+AC}$$ $$DC = BC\cdot\frac{AC}{AB+AC}$$
In $(1)$ , $$AD\cdot AD = AC\cdot AB-BC^2\cdot AB\cdot \frac{AC}{(AB+AC)^2}$$ $$AD^2 = AC\cdot AB\Bigl(1-\frac{BC^2}{(AB+AC)^2}\Bigr)$$ If $AB = c , BC = a , AC = b$: $$AD^2 = bc\Bigl( 1-\frac{a^2}{(b+c)^2}\Bigr)$$ Hence proved.
Writing $a := |BC|$, $b := |CA|$, $c := |AB| = \sqrt{a^2+b^2}$, and $r = |PE| = |PF|$ (so that $|PD| = 3r$), we have $$\begin{align} |\triangle ABC| &= |\triangle ABP| + |\triangle BCP| + |\triangle CAP| \\[4pt] \implies \qquad \frac{1}{2} |BC||CA| &= \frac{1}{2} \left(\; |AB| |PF| + |BC||PD| + |CA||PE| \;\right) \\[4pt] \implies \qquad a b &= c r + 3 a r + b r = r ( 3 a + b + c )\\[6pt] \implies \qquad r &= \frac{ab}{3 a + b + c} = \frac{ab}{3 a + b + \sqrt{a^2+b^2}} \end{align}$$
To address @DanielV's suggestion of generalizing to higher dimensions, consider a right-corner tetrahedron $OABC$, with right corner at $O$ and edge lengths $a := |OA|$, $b := |OB|$, $c := |OC|$. (Note that I'm changing notation slightly from the above.) Let a sphere with center $P$ and radius $r$ be tangent to the faces around vertex $A$, and let a congruent sphere (tangent to the first) be tangent to the faces around vertex $O$. Then $P$ has distance $r$ from faces $\triangle OAB$, $\triangle OCA$, $\triangle ABC$ (the ones touching $A$), and distance $3r$ from face $\triangle OBC$ (the one opposite $A$).
Here's a poor attempt at a diagram:
(In this case, the altitudes from $P$ are color-coded to match their parallel counterparts through $O$. The black altitude is to face $\triangle ABC$.)
Thus,
$$\begin{align} |OABC| &= |OABP| + |OBCP| + |OCAP| + |ABCP| \\[4pt] \implies \qquad \frac{1}{6}a b c &= \frac{1}{3}\left(\; r\;|\triangle OAB| + r \;|\triangle OCA| + r\;|\triangle ABC| + 3r\;|\triangle OBC| \;\right) \\[4pt] &= \frac{1}{3}r \cdot \frac{1}{2} \left(\; a b + c a + 3 b c + 2\;|\triangle ABC| \;\right) \\[6pt] \implies \qquad r &= \frac{abc}{3bc + ab + ca + 2\;|\triangle ABC|} \qquad (\star) \end{align}$$
Fun fact: The Pythagorean Theorem for Right-Corner Tetrahedra says that $$|\triangle ABC|^2 = |\triangle OBC|^2 + |\triangle OCA|^2 + |\triangle OAB|^2$$ so that we have $$|\triangle ABC| = \frac{1}{2} \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}$$ and $(\star)$ becomes $$r = \frac{abc}{3bc + ab + ca + \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}}$$
In $4$-dimensional space (where there's an analogous Pythagorean Theorem, as there is in any-dimensional space), we have $$r = \frac{abcd}{3bcd + acd + abd + abc + \sqrt{\;b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2\;}}$$ and so forth.
Incidentally, the matching-notation version of the initial answer is $$r = \frac{ab}{3b + a + \sqrt{\;b^2 + a^2\;}}$$
Best Answer
An elemenraty solution: In the following figure $|BC|=a,|CA|=b,|AB|=c$ and $[CD]$ angle bisector. Let's draw square $CEDF$ and $|CE|=x$. So, $|BE|=a-x$ and $|CD|=x\sqrt2$. Now $\triangle ABC \sim \triangle DBE$ and $$\dfrac{b}{a}=\dfrac{x}{a-x} $$
Therefore $x=\dfrac{ab}{a+b}$ and $|CD|=\dfrac{ab\sqrt2}{a+b}.$