In the triangle ABC, ab = AC, angle BAC = 60 degrees. D is a point outside the triangle (under BC), connecting ad, BD, CD. The angle BDC = 120 degrees How to find the relationship between angle BDA and angle CDA?

From known: triangle ABC is equilateral triangle
If ABC is used as a circle, the radians AC and ab are equal to 120 degrees,
Because the angle BDC is equal to 120 degrees = 1 / 2 (AB + AC) radian,
So point d must be on the circle,
So the angle BDA = 1 / 2 arc AB = 60 degrees, angle CDA = 1 / 2 arc AC = 60 degrees,
So angle BDA = angle CDA

The triangle ABC is an equilateral triangle, D is a point outside BC, and the angle BDC is 120 degrees, indicating that ad = BD + CD

Because the triangle ABC is a regular triangle, D is a point outside BC, and the angle BDC is 120 degrees
Then D is on the vertical bisector of BC side, that is, ad is the bisector of angle A
Another angle DCA = DBA = 90 ° CD = BD = 1 / 2ad
AD=BD+CD

The triangle ABC is an equilateral triangle, D is a point outside ABC, and the angle BDC is 120 degrees. It is proved that BD + CD = ad

prove:
Extend BD to point E so that DE=DC,
∵ BDC = 120 degrees, so ∵ CDE = 60 degrees
The △ CDE is an equilateral triangle
﹤ ECD = 60 degrees, CD = CE
∵ BCE = ACD, and △ ABC is an equilateral triangle, AC = BC,
∴ACD≌△BCE
∴AD=BE=BD+DE=BD+DC

In the equilateral triangle ABC, D is the outer point of the triangle ABC, and the angle BDC = 120 degrees, connecting ad

In the 20 degree triangle ABC, ab = AC, angle a = 20 degrees, ad = BC, calculate the degree of angle BDC

In ABC and ead of triangle, ab = EA, angle ABC = angle ead = 80 degrees, BC = ad (SAS) two triangles are congruent. The results show that the angle EDA = angle ACB = 80 degrees, angle BAC = angle AED, then angle Dec = 8

In the triangle ABC, the angle a = 20 degrees, ab= AC.AB There is a point on D, ad = BC, find: what is the angle of angle BDC? This problem requires very peculiar thinking

Let, angle BDC = X
DC/sin80=BC/sinx.(1)
DC/sin20=AD/sin(x-20)=BC/sin(x-20).(2)
(1)/(2)
sin(x-20)/sinx=sin20/sin80=2sin10cos10/cos10=2sin10
sin(x-20)=2sin10sinx
sinxcos20-cosxsin20=2sinxsin10
sinx(cos20-2sin10)=cosxsin20
tanx=sin20/(cos20-2sin10)=sin20/(sin70-sin10-sin10)
=sin20/(2sin30cos40-sin10)
=Sin20 / (sin50-sin10) = sin20 / 2sin20cos30 = radical 3 / 3
X = 30 degrees
That is, the angle BDC = 30 degrees

In the triangle ABC, the angle a = 90 degrees BD bisection angle ABC ad = 6cm BC = 15cm find the area D of BDC on AC

As de ⊥ BC
In RT △ abd and RT △ EBD
∵ ∠A=∠BED=90°
BD = BD (common side)
∴Rt△ABD≌Rt△EBD（H.L）
Then ad = ed = 6
 s △ BCD = 1 / 2 * BC * de = 45 square centimeter

As shown in the figure, in triangle ABC, angle a = 90 ° BD bisect triangle ABC, ad = 6cm, BC = 15cm, and calculate the area of triangle BDC

The tips are as follows
De is made by D and BC is perpendicular to E
Angle a = 90 degrees, angle DEB = 90 degrees, angle DBE = angle DBA, BD = BD
Triangle abd congruent triangle EBD
DE＝AD
S（BDC)＝BC*DE/2=15*6/2=45

As shown in the figure, in the triangle ABC, ∠ a = 90 °, BD bisection ∠ ABC, ad = 6, BC = 16, CE ⊥ BC, calculate the area s △ BDC

BD bisection ∠ ABC, ∠ a = 90 ° CE ⊥ BC, so de = Da = 6, (the distance from the point on the bisector to both sides of the corner is equal)
S△bdc =DE*BC/2=6*16/2=48

Given the triangle ABC, ab = AC, ∠ a = 20 ° and taking a point D on AB to make ad = BC, what degree is ∠ BDC equal to

On the inner side of △ ABC, the length of BC is taken as the equilateral △ BCE, and AE is connected, then AE is the vertical line of BC, and AE is also the angular bisector of ∠ A. in △ AEC, ∠ EAC = 10 °

In the triangle ABC, ab = AC, angle BAC = 60 degrees. D is a point outside the triangle (under BC), connecting ad, BD, CD. The angle BDC = 120 degrees How to find the relationship between angle BDA and angle CDA?