21.「2026广东揭阳惠来期末，★☆」(8分)如图,在$△ ABC$中,点D是边AB上一点,点M是边AC的中点,连接DM并延长至点N,使得$MN=DM$,连接AN,CN,CD,且$∠ AMD=2∠ MCD$.
(1)求证:四边形ADCN是矩形.
(2)若$∠ BAC=60°$,$BD=2AD=4$,求点D到边BC的距离.

解析 (1)证明:$\because$ 点$M$是边$AC$的中点,$\therefore AM=CM$,
$\because MN=DM$,$\therefore$ 四边形$ADCN$是平行四边形.
$\because ∠ AMD=∠ MCD+∠ MDC$,$∠ AMD=2∠ MCD$,
$\therefore ∠ MCD=∠ MDC$,$\therefore DM=CM$,
$\therefore AM=CM=MN=DM$,$\therefore AC=DN$,
$\therefore$ 平行四边形$ADCN$是矩形.$·$ (4分)
(2)如图,过点$D$作$DE⊥ BC$于点$E$,

$\because BD=2AD=4$,$\therefore AD=2$,
由(1)可知,四边形$ADCN$是矩形,$\therefore ∠ ADC=90°$,
$\because ∠ BAC=60°$,$\therefore ∠ ACD=90°-60°=30°$,
$\therefore AC=2AD=4$,$\therefore CD=\sqrt{AC^2-AD^2}=\sqrt{4^2-2^2}=2\sqrt{3}$,
$\because ∠ BDC=180°-∠ ADC=90°$,
$\therefore BC=\sqrt{BD^2+CD^2}=\sqrt{4^2+(2\sqrt{3})^2}=2\sqrt{7}$,$·································$ (6分)
$\because DE⊥ BC$,$\therefore S_{△ BCD}=\dfrac{1}{2}BC· DE=\dfrac{1}{2}BD· CD$,
$\therefore DE=\dfrac{BD· CD}{BC}=\dfrac{4×2\sqrt{3}}{2\sqrt{7}}=\dfrac{4\sqrt{21}}{7}$,即点$D$到边$BC$的距离为$\dfrac{4\sqrt{21}}{7}$. $·$ (8分)