Giải thích các bước giải:

a.Ta có: $\Delta ABC$ đều, $M$ là trung điểm $BC\to AM\perp BC$

                $AD\perp (ABC)\to AD\perp BC$

 $\to BC\perp (ADM)$

$\to BC\perp DM$

$\to BD\perp MH$

b.Ta có: $BC\perp (ADM)\to BC\perp AH$

                $AH\perp DM$

$\to AH\perp (BCD)$

c.Gọi $E$ là trung điểm $AB$

$\to ME//AC, EM=\dfrac12AC=\dfrac12a$

Ta có: $DE=\sqrt{AD^2+AE^2}=\sqrt{(\dfrac45a)^2+(\dfrac12a)^2}=\dfrac{a\sqrt{89}}{10}$

            $AM=\dfrac{a\sqrt3}2\to DM=\sqrt{DA^2+AM^2}=\dfrac{a\sqrt{139}}{10}$

$\to \cos\widehat{DME}=\dfrac{MD^2+ME^2-DE^2}{2\cdot MD\cdot ME}$

$\to \cos\widehat{DME}=\dfrac{15\sqrt{139}}{278}$

$\to\widehat{DME}=\arccos(\dfrac{15\sqrt{139}}{278})$

Vì $ME//AC\to \widehat{AC, DM}=\widehat{DME}=\arccos(\dfrac{15\sqrt{139}}{278})$