In known quadrilateral ABCD, angle abd = 12 °, angle DBC = 36 °, angle ACB = 48 ° and angle ACD = 24 °, try to find the degree of angle ADB I can draw pictures according to the title,

In known quadrilateral ABCD, angle abd = 12 °, angle DBC = 36 °, angle ACB = 48 ° and angle ACD = 24 °, try to find the degree of angle ADB I can draw pictures according to the title,

Let's make an equilateral triangle Abe (AE intersects BC) with ab as the edge, connect CE, ab = be = AE, angle abd = 12 °, angle DBC = 36 ° and angle ACB = 48 °, then angle BAC = 84 °, angle ABC = angle ACB, ab = AC = AE, angle AEC = angle ace, angle CAE = 84-60 = 24 °, angle ace = (180-24) / 2 = 78 °, angle BCE = 78-48 = 30 ° angle DB

Known: as shown in the figure, ∠ abd = ∠ DBC, ∠ ACD = ∠ DCE. (1) if ∠ a = 50 °, find the degree of ∠ D; (2) guess the relationship between ∠ D and ∠ a, and explain the reason; (3) if CD ‖ AB, judge the relationship between ∠ ABC and ∠ a

(1) For △ BCD, ∠ DCE = ∠ DBC + ∠ D, and ∵ ∵ abd = ∠ DBC, ∠ ACD = ∠ DCE, ∵ ACD = ∠ abd + ∠ D. from the properties of the outer angle of the triangle, ∠ a + ∠ abd = ∠ D + ∠ ACD can be solved by the above two formulas, ∠ d = 12 ∠ a = 25 °; (2) from (1), ∠ d = 12 ∠ a; (3) if CD ‖ a

As shown in the figure, in the convex quadrilateral ABCD, ∠ abd = 16 °, ∠ DBC = 58 ° and ∠ ACD = 30 °, calculate the degree of ADB

Less condition? Is it ∠ BCD = 90

It is known that in ABC, EF is the midpoint CD bisection angle BCA of AB and AC respectively. It is proved that ad is perpendicular to DC

∵ E and F are the midpoint of AB and AC respectively
∴EF∥BC
∴∠FDC=∠BCD
CD bisection BCA
∴∠FDC=∠FCD
∴DF=FC=FA
∴∠ADC=90°
That is ad ⊥ DC

It is known that in △ ABC, e and F are the midpoint of AB and AC respectively, CD bisects ∠ BCS intersects EF with D, and ad ⊥ DC is proved

It is proved that E and F are the midpoint of AB and AC respectively
So, EF / / BC
Therefore, FDC = BCD
Because CD bisects and BCA intersects EF with D
So, FCD = BCD
Therefore, ∠ FDC = ∠ FCD
So EF = FC
And F is the midpoint of AC, AF = FC
That is, EF = FC = AF
So △ ADC is a right triangle, ∠ ADC = 90 degrees
That is ad ⊥ DC

It is known that in the oblique triangle ABC, the intersection of the heights on both sides of AB and AC is h, and the degree of BHC is calculated

Let AB be a high intersection of AB and D
AC and e
For the same vertex angle, so ∠ BHC = ∠ DHE
Because h is the intersection of high, ADC = ∠ AEB = 90 degree
Therefore, a + DHE = 180 degree
Therefore, BHC = DHC = 125 degree

It is known that, as shown in the figure, there are two points D and E in △ ABC

It is known that, as shown in the figure, in △ ABC, ∠ ABC = 66 ° and ∠ ACB = 54 ° be and CF are the heights of AC and AB on both sides, and they intersect at point h. calculate the degree of ∠ Abe and ∠ BHC

In ∵ △ ABC, ∵ ABC = 66 °, ∵ ACB = 54 °, ∵ a = 180 ° - ∵ ABC - ∵ ACB = 180 ° - 66 ° - 54 ° = 60 °, ∵ be ⊥ AC, ∵ AEB = 90 °, ∵ Abe = 90 ° - a = 90 ° - 60 ° = 30 °; similarly, ∵ CF ⊥ AB, ∵ BFC = 90 °, ∵ BHF = 90 ° - Abe = 90 ° - 30 ° = 60 °, ∵ BHC = 180 ° - BHF = 180 ° - 60 ° = 120 °

As shown in the figure, in the triangle ABC, angle A: angle ABC: angle ACB = 3:4:5, BD and CE are the heights of the sides AC and ab respectively, BF and CE intersect at point h, and the angle BHC is obtained

∵∠A:∠ABC:∠ACB=3:4:5
∴∠A=3*180/(3+4+5)=45°
∵CE⊥AB,BD⊥AC
∴∠AEC=90° ∠ADB=90°
∴∠EHD=360-90-90-45=135°
∵∠BHC=∠DHE
∴∠BHC=135°

In the triangle ABC, ∠ a = 60 °, BD and CE are the heights of AC and ab respectively. H is the intersection of BD and CE to find the degree of ∠ BHC
RT. very urgent, eh

Because ∠ a = 60 ° and BD vertical AC, then ∠ ADB = 90 °
So, abd = 30
Similarly, ACB = 30
So angle bhe = 60, angle CHD = 60
In aehd, angle EHD = 360-90-90-60 = 120
Another angle BHC is the diagonal of EHD
So angle BHC = angle EHD = 120