Question Number 221791 by hardmath last updated on 10/Jun/25
Commented by hardmath last updated on 10/Jun/25
$$\mathrm{ABC}\:-\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{MK}\:\parallel\:\mathrm{AB} \\ $$$$\mathrm{MN}\:\parallel\:\mathrm{AC} \\ $$$$\mathrm{ML}\:\bot\:\mathrm{AB} \\ $$$$\mathrm{AC}\:=\:\mathrm{a} \\ $$$$\mathrm{MN}\:=\:\mathrm{x} \\ $$$$\mathrm{MK}\:=\:\mathrm{y} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{ML}\:=\:\frac{\left(\mathrm{a}\:-\:\mathrm{x}\:-\:\mathrm{y}\right)\centerdot\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$