Circles and Triangles

Notice that $\angle ACD=\angle ABD$ because they are inscribed angles subtended by the same chord $AD$ . The same occurs with $\angle CAB$ and $\angle CDB$ : they are equal because they are inscribed angles subtended by the same chord $CB$ . It means that $\boxed{\triangle AIC \sim \triangle BID}$ .

The segment $HI$ is perpendicular to the segment $AC$ , then $\triangle AIC \sim \triangle AIH \sim \triangle CHI$ . Using this information we can deduce that $\angle AIH= \angle ACD = \angle BIG = \angle ABD$ , which means that $\triangle BIG$ is isosceles and $IG=BG=19.5$

Using the same method as above we can deduce that $\angle CIH= \angle CAB = \angle GID = \angle CDB$ , which means that $\triangle GID$ is isosceles and $IG=GD=19.5$ .

$BD=BG+GD=19.5+19.5=39$ . By Pythagorean Theorem $39^2-15^2=ID^2 \Rightarrow 1521-225=ID^2 \Rightarrow ID=36$ . By the Intersecting Chords Theorem $CI\cdot ID=AI \cdot IB$ . Then, $CI\cdot 36=24 \cdot 15 \Rightarrow 36 \cdot CI =360 \Rightarrow CI=10$ .

If $CI=10$ and $AI=24$ , $AC^2=10^2+24^2=100+576=676\Rightarrow AC=26$ . Finally, by the triangles similarities we have that $\frac { CI }{ AC } =\frac { HI }{ AI }$ . Substituting with the values we have, $\frac { 10 }{ 26 } =\frac { HI }{ 24 } \Rightarrow 26\cdot HI=240$ . Then $\boxed{HI\approx 9.23}$

Drawing $\alpha$ and $\beta$ angles such that their sum equals $90^{\circ}$ (for example $\angle BAC = \alpha$ and $\angle DCA = \beta$ ) is easy to realize that $\bigtriangleup AIH \sim \bigtriangleup DBI$ . Also, we find $\angle DIG = \angle IDG = \alpha$ and $\angle GIB = \angle IBG = \beta$ , so $\overline{IG} \cong \overline{BG} \cong \overline{DG}$ .

Using this information we can set our proportionality equation: $\frac{x}{\overline{AI}}=\frac{\overline{IB}}{2\cdot \overline{IG}}$ $\Rightarrow x \approx 9,23$