Ta có:
$AHCK$ là hình chữ nhật (Vì $\widehat{H}=\widehat{C}=\widehat{K}=90^o$)
$→ AK=HC=12$ (cm)
$AH^2=HB.HC=12.3=36 → AH=6$ (cm)
$AC=\sqrt[]{AH^2+HC^2}=\sqrt[]{6^2+12^2}=6\sqrt[]{5}$ (cm)
$\dfrac{1}{AC^2}+\dfrac{1}{AI^2}=\dfrac{1}{AK^2}$
$↔ \dfrac{1}{AI^2}=\dfrac{1}{AK^2}-\dfrac{1}{AC^2}=\dfrac{1}{720}$
$→ AI=\sqrt[]{720}=12\sqrt[]{5}$ (cm)
$CI=\sqrt[]{AC^2+AI^2}=\sqrt[]{(6\sqrt[]{5})^2+(12\sqrt[]{5})^2}=30$ (cm)
$CB=BH+HC=3+12=15$ (cm)
$IK=\sqrt[]{AI^2-AK^2}=\sqrt[]{(12\sqrt[]{5})^2-12^2}=24$ (cm)
$BH=3$ (cm)
$→ \dfrac{CI^3}{CB^3}=\dfrac{30^3}{15^3}=8$
$\dfrac{IK}{BH}=\dfrac{24}{3}=8$
Vậy $\dfrac{CI^3}{CB^3}=\dfrac{IK}{BH}$ (điều phải chứng minh)
