In the triangle ABC, ab = AC, angle a = 120 degrees, D is the midpoint of BC, De is perpendicular to AB and E, and the ratio of AE to EB is calculated

In the triangle ABC, ab = AC, angle a = 120 degrees, D is the midpoint of BC, De is perpendicular to AB and E, and the ratio of AE to EB is calculated

If ∠ bad = 60 ° is known, then ∠ ade = ∠ abd = 30 °
Let ad = a, then AE = A / 2, ab = 2A, EB = 2a-a / 2 = 3A / 2,
So AE: EB = A / 2:3a / 2 = 1:3

As shown in the figure, in △ ABC, ∠ a = 120 °, D is the midpoint of BC, de ⊥ AB is in E, and AE: EB is obtained

There should be another condition: ab = AC?
Because AAB = AC, ∠ a = 120 °, so, ∠ B = 30 °, bad = 1 / 2, BAC = 60,
AB=2AD.
Because De is perpendicular to AB, so ∠ ade = 30 ° ad = 2ae,
So AB = 2ad = 4ae, be = ab-ae = 3aE,
So AE: be = 1:3

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 °. D is the midpoint of BC, de ⊥ AB is proved at point E: EB = 3EA

It is proved that: ∵ AB = AC, ∵ BAC = 120 ∵ B = ∵ C = 30 ∵ D is the midpoint of BC ∵ ad ⊥ BC and ad bisects ∵ BAC, ∵ bad = 60 ∵ ADB = 90 ∵ ad = 12ab and