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Giải:
Tự vẽ hình nhé! 
Ta xét $\Delta$ ACD.
Ta có:
$cos \widehat{CDA} = \frac{CD^2 + AD^2 - AC^2}{2CD.AD}$
$cos \widehat{DAC} = \frac{AD^2 + AC^2 - CD^2}{2AD.AC}$
$cos \widehat{ACD} = \frac{AC^2 + CD^2 - AD^2}{2AC.CD}$
Từ gt cho \Rightarrow $CD^2 + AD^2 - AC^2 = AD^2 + AC^2 - CD^2 = AC^2 + CD^2 - AD^2$
\Rightarrow $cos \widehat{CDA}, cos \widehat{DAC}, cos \widehat{ACD} $ > 0.
\Rightarrow $\widehat{CDA}, \widehat{DAC},\widehat{ACD}$ < $90^o$
Vậy: $\Delta$ ACD có cả 3 góc nhọn. (Đpcm)
.