Find $\overline{CD}$ in this diagram

The inradius of $\triangle ABC$ is $r = \dfrac{2A}{P} = \dfrac{(3)(4)}{3 + 4 + 5} = 1$ . Let $x = \overline{CD}$ , then using the same formula,

$1 = \dfrac{2A}{P} = \dfrac{(x)(4) } {5 + x + \sqrt{4^2 + (3 + x)^2} }$

Hence,

$4 x= 5 + x + \sqrt{x^2 + 6 x + 25 }$

from which,

$\sqrt{x^2 + 6 x + 25 } = 3 x - 5$

This equation solves easily to $x = \dfrac{9}{2} = \boxed{4.5}$

Let the radius of the incircles be $r$ . From the relation $[\triangle ABC]=r\cdot s_1$ , we have, $r=\frac{2\triangle_1}{AB+BC+CA}=\frac{12}{12}=1$ Similarly, the inradius of $\triangle BCD$ is also $r$ . Let the length of side $CD=x$ be $x$ and $BD=y$ . Then, $r=\frac{[\triangle BCD]}{s_2}=\frac{2\triangle_2}{CB+BD+DC}$ $1=\frac{4x}{5+x+y}$ $\implies y=3x-5 \; \; \; \; \; \; \; (1)$ Also, note that $\triangle ABD$ is right angled at $A$ . By the Pythagorean theorem, $AB^2+AD^2=BD^2$ $4^2+(3+x)^2=y^2 \implies 25+6x+x^2=y^2 \; \; \; \; \; \; \; (2)$ Combing equations $(1)$ and $(2)$ , $25+6x+x^2=(3x-5)^2=9x^2-30x+25$ $\implies 8x^2=36x\implies x=\frac{9}{2}$ Therefore, the desired length of $\boxed{CD=4.5}$

Let the radius of the incircles be $r$ . Then the area of $\triangle ABC$ , $[ABC] = \dfrac {3+4+5}2r = \dfrac {3 \times 4}2 \implies 6r = 6 \implies r = 1$ .

Let $CD=x$ and $BD=y$ . By Pythagorean theorem , $y = \sqrt{(x+3)^2 + 4^2} = \sqrt{x^2+6x+25}$ . And the area of $\triangle ABD$ :

$\begin{aligned} [ABD] & = [ABC]+[BCD] \\ \frac {4(x+3)}2 & = \frac {4\times 3}2 + \frac {5+x+y}2 r \\ 2x + 6 & = 6 + \frac {5+x+y}2 \\ 4x & = 5 + x + y \\ 3x - 5 & = y = \sqrt{x^2+6x+25} & \small \blue{\text{Squaring both sides}} \\ 9x^2 - 30x + 25 & = x^2 + 6x + 25 \\ 8x^2 - 36x & = 0 \\ x(2x - 9) & = 0 & \small \blue{\text{Since }x>0} \\ \implies x & = \frac 92 = \boxed{4.5} \end{aligned}$

line EF intersects BC at G, extend it intersecting AB at G. And circle F tangents AD, BC and BD at P, Q, R respectively. Then:

$r = \dfrac {2S}{C} = \dfrac {2 \cdot \frac 12 \cdot 3 \cdot 4}{3+4+5} = 1 \\ HE : EG = BH : BG = BA : BC \\ = 4 : 5 \\ \implies EG = \dfrac 54, GF = EG = \dfrac 54 \\ AP = HF = 1 + 2 \times \dfrac 54 = \dfrac 72 \\ \implies CQ = CP = AP - AC = \dfrac 12 \\ \implies BR = BQ = BC - CQ = \dfrac 92 \\ AB^2 = BD^2 - AD^2 \\ = (BD + AD)(BD-AD) \\ = (BR+DR+AP+PD)(BR+DR-AP-DP) \\ =(BR+AP+2PD)(BR-AP) \\ 4^2 = (\dfrac 92 + \dfrac 72 + 2PD)(\dfrac 92 - \dfrac 72)\\ PD = 4 \\ \implies CD = CP + PD = \dfrac 12 + 4 = \boxed{\dfrac 92}$