`\color{#FF0000}{T}\color{#FF7F24}{i}\color{#FFFF00}{t}\color{#7FFF00}{u}\color{#00F5FF}{u}\color{#FF1493}{u}`

$1)$ Xét $\triangle$ $ADH$ và $\triangle$ $AHB$ có:

Chung `\hat{DAH}`

`\hat{ADH}=\hat{AHB}=90^o`

`->` $\triangle$ $ADH$ $\backsim$ $\triangle$ $AHB$ $(G-G)$

`-> (AD)/(AH)=(AH)/(AB)`

`-> AD.AB=AH^2 (1)`

Tương tự `(1)` ta có: `AE.AC=AH^2`

`-> AD.AB=AE.AC`

$2)$ Tương tự `(1)` ta có: `BH^2=BD.BA`

`-> (AH^2)/(BH^2)=(AD.AB)/(BD.BA)=(AD)/(BD)`