As shown in the figure, in △ ABC, D is a point on the edge of BC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ BAC = 63 ° to find the degree of angle DAC

As shown in the figure, in △ ABC, D is a point on the edge of BC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ BAC = 63 ° to find the degree of angle DAC

∠4=∠1+∠2
Because ∠ 1 = ∠ 2
So ∠ 4 = 2 ∠ 1 = 2 ∠ 2
In the same way, 3 = 2 1 = 2 2 2
And ∠ 3 + ∠ 4 + ∠ DAC = 180 °
That is, 2 ∠ 1 + 2 ∠ 1 + ∠ DAC = 180 degrees
As shown in the figure
∠1+∠DAC=∠BAC=63°∠DAC=63°-∠1
Therefore, 3 ∠ 1 + 63 ° = 180 °
The solution is ∠ 1 = 39 °
The results show that ∠ DAC = 63 ° - ∠ 1 = 24 °

As shown in the figure, in △ ABC, ∠ ACB = 90 °, AC = BC, P is a point in the triangle, and PA = 3, Pb = 1, PC = 2, calculate the degree of ∠ BPC Here's the picture.  

The simple solution takes C as the rotation center and rotates the △ cap by 90 degrees,
If point a and point B coincide, then p → Q
CQ=CP,BQ=AP,∠PCQ=90°.
The △ PCQ is an isosceles right triangle,
PQ^2=4+4=8,
And ∵ PQ ^ 2 + Pb ^ 2 = 8 + 1 = 9 = BQ ^ 2
∴∠BPQ=90°,
Therefore, BPC = ∠ bpq + ∠ CPQ = 90 ° + 45 ° = 135 °

As shown in the figure, in △ ABC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ BAC = 54 ° to find the degree of ∠ DAC

∵∠1=∠2，∠3=∠4，
∴∠4=2∠1=2∠2=∠3
∴∠2+∠3=3∠2=126°
∴∠2=∠1=42°  
∴∠DAC=54-42=12°．

As shown in the figure, in △ ABC, D is a point on BC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ BAC = 120 °, then ∠ DAC=______ Degree

∵∠BAC=120°，
∴∠2+∠3=60°①
∵∠1=∠2，
∴∠4=∠3=∠1+∠2=2∠2②
Substituting (2) into (1) gives: 3 ∠ 2 = 60 °,
∠2=20°，
∴∠1=20°．
∴∠DAC=120°-20°=100°．
So the answer is: 100

As shown in the figure, in triangle ABC, D is a point on the edge of BC, angle 1 = angle 2, angle 3 = angle 4, angle BAC = 63 ° and find the degree of angle DAC

Angle 1 = angle 2, angle 3 = angle 4
Angle 3 = angle 1 + angle 2 = 2 * angle 1 = angle 4
Angle BAC + angle 2 + angle 4 = 180 degrees
63 degree + angle 1 + 2 * angle 1 = 180 degree
3 * angle 1 = 180 ° - 63 ° = 117 °
Angle 1 = 117 ° / 3 = 39 °
Angle DAC = angle BAC - angle 1 = 63 ° - 39 ° = 24 °

As shown in the figure, in the triangle ABC, angle ABC = angle c = 2 angle a, BD is perpendicular to AC to D, and the degree of angle DBC is obtained

∵∠ABC＝∠C＝2∠A,∠A+∠ABC+∠C＝180
∴∠A+2∠A+2∠A＝180
∴∠A＝36
∴∠C＝2∠A＝72
∵BD⊥AC
∴∠DBC+∠C＝90
∴∠DBC＝90-∠C＝18°
The math group answered your question,

In the triangle ABC, the angle c = angle ABC = 2 angle a, BD is the height on the edge of AC. calculate the degree of angle DBC

∠A+∠B+∠C=180°
∠A+2∠A+2∠A=180°
5∠A=180°
∠A=36°
∠B=∠C=72°
∠DBC=90°-∠C=90°-72°=18°

As shown in the graph, in △ ABC, angle c = angle ABC = 2 angle a, BD is the height of AC edge

The process is as follows

As shown in the figure, it is known that in the triangle ABC, ab = AC, point D is a point on edge AC, and angle a = 2, angle DBC. It is proved that BD is perpendicular to AC

Let DBC be x, then angle a is 2x. Because AC = AB, so angle c = 90-x. so angle BDC is 90 degrees

As shown in the figure, ad ‖ BC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ad = 4, BC = 2, then ab=______ .

Prolonging ad, be intersects with F. ∵ ad ≓ BC,  DAB + ∵ ABC = 180 °, ∵∵ 1 = ∵ 2, ∵ 3 = ∠ 3, ∵ Abe = 90 °, that is, AE ⊥ be, ∵ ad ∥ BC, ∵ 4 = ∠ f = ∠ 3, ∵ AB = AF, ∵ be = EF, ad ∥ BC, ∵ CE = De, BC = DF, ﹤ AF = AD + DF = AD + BC = 6, ab = AF = 6