In the triangle ABC, the bisector angle of Bo ABC ad is vertical to BD, and the perpendicular foot is d AE = EC to prove the parallel BC of de

In the triangle ABC, the bisector angle of Bo ABC ad is vertical to BD, and the perpendicular foot is d AE = EC to prove the parallel BC of de

Prolonging ad to f
∵∠ ABO = ∠ FBO (Bo bisector angle ABC)
Bo = Bo (common side)
∠ADB=∠FDB=90°
≌△ ADB ≌ △ FDB (two corners clamped)
That is, D is the midpoint of AF
And ∵ AE = EC, that is, e is the midpoint of AC
∴DE‖BC

As shown in the figure, AE is the bisector of ∠ CAD at the outer angle of △ ABC, and the extension line of AE intersecting BC is at point E, ∠ B = 30 ° and ∠ DAE = 62 °, then ∠ ade=--------

As shown in the figure, point D can rotate around point e along with edge ed (with the ad angle unchanged). Therefore, the missing condition is known. Please show me the original picture

There is an isosceles triangle ABC. The height of the side BC is exactly equal to half of the side BC
RT

If BC is the bottom, then BAC is an isosceles right triangle
Angle BAC = 90
If BC is waist
The height and the other waist form a right triangle, and the hypotenuse is twice of the right side,
Apex angle 30
Angle BAC = 75

As shown in the figure, in △ ABC, be and CF are respectively the high lines on the sides of AC and ab. be and CF intersect at O, connect Ao, intersect BC at D, and △ BCF ≌ △ CBE, ∠ ABC = 70 ° to find the degree of ∠ 1 and ∠ 2

∵ △ BCF ≌ △ CBE, ≌ ∠ FBC = ∠ ECB = 70 °, ∩ BAC = 180 ° - ∠ FBC - ∠ ECB = 40 °, ab = AC, ≌ be, CF are the high lines on the sides of AC and AB, be and CF intersect at O, ∵ ad ⊥ BC, ∩ 1 = ∠ 2 = 12 ∠ BAC = 20 °