In RT triangle ABC, angle ACB is equal to 90 degrees, angle a = 30 degrees, BD is the bisector of triangle ABC, De is perpendicular to point E 1, In RT triangle ABC, angle ACB is equal to 90 degrees, angle a = 30 degrees, BD is the bisector of triangle ABC, De is perpendicular to point E Q: connect EC and prove that the triangle EBC is an equilateral triangle

In RT triangle ABC, angle ACB is equal to 90 degrees, angle a = 30 degrees, BD is the bisector of triangle ABC, De is perpendicular to point E 1, In RT triangle ABC, angle ACB is equal to 90 degrees, angle a = 30 degrees, BD is the bisector of triangle ABC, De is perpendicular to point E Q: connect EC and prove that the triangle EBC is an equilateral triangle

It is proved that ABC = 60 degrees
Because BD is the triangle ABC bisector
So angle abd = angle CBD = 30 degrees
Because angle a = 30 degrees
Ad = BD
Because De is perpendicular to ab at point E
All △ ade = △ BDE
ae=be
Because △ ABC City RT △
AB = 2BC
BC = be
Because the angle ABC = 60 degrees
The triangle EBC is an equilateral triangle
You can look at that step without subtracting it

As shown in the figure, it is known that in △ ABC, ∠ C = 90 ° and D and E are on BC; and BD = de = EC = AC (1) point out the similar triangles in the figure and prove your conclusion; (2) find the value of ∠ B + ∠ ADC + ∠ AEC

(1) Reason: in △ AED and △ BEA, ∵ ABC, ∵ C = 90 °, BD = de = EC = AC, ∵ AEC is isosceles right triangle, be = BD + de = 2bd = 2Ac, ∵ AEC = 45 °, that is sin ∵ AEC = ACAE, ∵ AE = AC22 = 2Ac, ∵ Aede = beae = 22 = 2, ∵ AED = ∵ BEA

As shown in the figure, ⊙ o with BC as the diameter intersects AB at point D, AC at point E, and BD = EC in △ ABC. Try to judge the shape of △ ABC and prove it

In △ BOD and △ Coe, OD = oeob = ocbd = CE, ≌ BOD ≌ COE (SSS), ≌ B = C, ≌ AB = AC, ≌ ABC are isosceles triangles

AB is the diameter of the circle O, BC is the chord, angle ABC = 30 degrees, through the center of the circle O as OD, vertical BC, intersecting arc BC at point D, connecting DC. Determine the shape of the quadrilateral acdo, write the proof

Acdo is rhombic, which is proved as follows:
∵ AB is the diameter of circle O and BC is the chord
∴∠ACB=90°
Also: ∠ ABC = 30
∴AC=1/2AB=AO=OC
The AOC is an equilateral triangle
∴∠AOC=60°
Also: OD ⊥ BC
∴OD∥AC
∴∠BOD=∠OAC=60°
∴∠COD=180°-∠AOC-∠BOD = 60°
And: OC = OD
The ∧ OCD is an equilateral triangle
∴CD=OC=OD
∴OA=AC=CD=DO
The acdo is a diamond

Given that angles a, B and C are three internal angles of △ ABC, the equation x2 − xcosacosb − cos2c2 = 0 with a root of 1, then the shape of △ ABC is______ A triangle

∵ 1 is a root of the equation x2-xcosacosb-cos2c2 = 0, ∵ 1-cosacosb-1 + Cosc2 = 0, ∵ 12-cosc2 = cosacosb = 12 [cos (a + b) + cos (a-b)] = - 12cosc + 12cos (a-b), ∵ 12cos (a-b) = 12, ∵ cos (a-b) = 1, and a and B are the internal angles of △ ABC, ∵ a = B. ∵ a