In RT △ ABC, ∠ C = 90 degree, ∠ a = 30 degree, then BC: AC: ab = what

∵∠ C = 90 degree, ∠ a = 30 degree (known)
In a right triangle, if there is an angle of 30 degrees, then the right side of the angle is equal to one of the hypotenuses
(half)
AB"=BC"+AC"
(2BC) "= BC" + AC "(equivalent substitution)
‖ AC "= radical 3bC (equality property)
And ∵ AB = 2BC (proved)
Ψ BC: AC: ab = 1: radical 3:2

As shown in the figure, cdae is the height of triangle ABC, angle B = 45 degrees, AC = 4, and De is obtained
== proof
It's easy to understand~

CD, AE are high ∠ BAE + ∠ B = ∠ B + ∠ BCD
∠BAE= ∠BCD
∠ B=∠B
Triangle BAE ∽ triangle BCD
AB:BC=BE:BD
∠B=∠B
Triangle BAC ∽ triangle bed
BA:BE=AC:DE
AE is high, B = 45 degree
AB=BE√2
AC:DE= √2
DE=AC/√2=4/√2=2√2

In the known RT triangle ABC, the angle ABC = 90 degrees, AC = BC, CE is perpendicular to E
The extension line of BF parallel AC intersection CE is at F, proving that ACD of triangle is equal to CBF of triangle
Wrong-
Is RT triangle ABC, angle ACB = 90 degrees. AC = BC, CE is perpendicular to ad, BF is parallel to AC, and the extension line of CE is at F.

It is proved that ∵ BF ∥ AC ∥ CBF + ∠ ACD = 180 ° (two straight lines are parallel, and the inner angles of the same side are complementary) ∵ ACD = 90 ∥ CBF = ∠ ACD = 90 ⊥ CE ⊥ ad ∥ CAE + ∠ ace = 90 ° and ∵ ACE + ∠ BCF = 90 ∥ CAE = ∠ BCF (the remaining angles of the same angle are equal) in △ ACD and △ CBF ∥ a

In △ ABC, it is known that ∠ cab = 60 °, D and E are points on edges AB and AC respectively, and ∠ AED = 60 °, ed + DB = CE, ∠ CDB = 2 ∠ CDE, then ∠ DCB = ()
A. 15°B. 20°C. 25°D. 30°

As a result, EC = de + BD = AB = BF, de = FC, and = choose 65125; 1 = 2 = 2 = 120 ° 8780; EDC ≌ FCB, and 8780; FCB, \≌8780;≌≌≌ ≌ FCB, \\8780;≌ \ \\\ \\\ \\≌\\\\\\\\\\\\\\\\it's not easy

In RT △ ABC, ∠ C = 90 degree, ∠ a = 30 degree, then BC: AC: ab = what