As shown in the figure, in the triangle ABC, the angle ACB = 90 °, CD is perpendicular to AB, the perpendicular foot is D, and the angle a = 60 ° to prove BD = 3aD

As shown in the figure, in the triangle ABC, the angle ACB = 90 °, CD is perpendicular to AB, the perpendicular foot is D, and the angle a = 60 ° to prove BD = 3aD

Because angle a = 60, angle c = 90, CD is perpendicular to ab
So AC = 2ad, ab = 2Ac, then AB = 4AD
So BD = ab-ad = 4ad-ad, that is BD = 3aD

As shown in the figure, in the triangle ABC, ∠ ACB = 90 degree, CD is vertical, AB is vertical, D ∠ a = 60, and BD = 3aD is proved
Please answer completely

In RT triangle ABC, ∠ ACB = 90, ∠ a = 60, so AB = 2Ac
In the RT triangle ACD, ∠ ADC = 90, ∠ a = 60, so AC = 2ad
So AB = 4AD, and because AD + BD = AB, BD = 3aD

As shown in the figure, in the triangle ABC, the angle ACB is equal to 90 degrees, D is any point on AC, De is perpendicular to AB and E.M, n is the midpoint of BD and CE respectively, and Mn is perpendicular
Yu CE

Prove: connect cm, EM, because the angle ACB = 90 degrees, so the triangle BCD is a right triangle, because m is the midpoint of BD, so cm is the middle line of the right triangle BCD, so cm = 1 / 2bd, because De is perpendicular to AB, so the angle bed = 90 degrees, so the triangle bed is a right triangle, so EM is the middle line of the right triangle bed, so EM = 1 / 2