Ta có: `9IC=7ID`

Mà: `IC+ID=CD`

`=>(IC)/(CD)=(IC)/(IC+ID)=(IC)/(IC+9/7IC)=7/16`

`(ID)/(CD)=1-(IC)/(CD)=1-7/16=9/16`

`triangle ABC` vuông tại `B` có:

`cot hat{BAC}=(AB)/(BC)=4/2=2`

Mà: `hat{BAC}+hat{CAD}=90^@`

Nên: `tan hat{CAD}=cot hat{BAC}=2`

`=>cos^2 hat{CAD}=1/(1+tan^2 hat{CAD})=1/(1+2^2)=1/5`

Mà: `hat{CAD}<90^@`

`=>cos hat{CAD}=1/(\sqrt{5})`

Áp dụng định lý Pytago `triangle ABC` vuông tại `B`:

`AC^2=AB^2+BC^2=4^2+2^2=20`

`=>AC+2\sqrt{5}`

Ta có: `vec{AI}=(IC)/(CD)vec{AD}+(ID)/(CD)vec{AC}`

`=7/16vec{AD}+9/16vec{AC}`

`=>AI^2=(7/16vec{AD}+9/16vec{AC})^2`

`=49/256AD^2+81/256AC^2+63/128*vec{AD}*vec{AC}`

`=49/256AD^2+81/256AC^2+63/128*AD*AC*cos hat{CAD}`

`=49/256*6^2+81/256*20+63/128*6*2\sqrt{5}*1/(\sqrt{5})`

`=153/8`

`=(3\sqrt{34})/4`

`=>a=3,b=4`

Vậy: `a+b=3+4=7`