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As shown in the figure, in the RT triangle ABC, ∠ = 90 ° ad bisect ∠ BAC, and ∠ B = 3 ∠ bad, find the degree of ∠ ADC
Let ∠ CAD = x, then ∠ cab = x (bisector) ∠ cab = x + x = 2x ∵ ? B = 3 ∠ bad ? B = 6x. According to the sum of inner angles of the triangle is 180 °∠ C + ∠ B + ∠ cab = 180 ° 90 + 6x + 2x = 180 ° x = 11.25 CAD = 11.25 ° according to the sum of triangle angles, the sum is 180 ° ADC = 180 ° - ∠ CAD - ∠ C = 180
As shown in Figure 1, in the triangle ABC, if angle B = 40 °, angle c = 60 ° and ad bisect angle BAC, then the degree of angle ADC is?
Because the angle B = 40 ° and the angle c = 60 ° according to the sum of the internal angles of the triangle is equal to 180 ° then the angle a is 80 ° and
Because ad bisects the angle BAC, so ∠ DAC = 40 degrees
Therefore, ADC = 180 ° - 40 ° - 60 ° = 80 °
As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC. The test shows that s △ abd: s △ ADC = AB: AC
It is suggested that de ⊥ AB, DF ⊥ AC and foot drop are e and f respectively ABD:S Δ ACD = (AB * de / 2): (AC * DF / 2) = AB: AC explain the reasons:
As shown in the figure, in triangle ABC, D is a point on the edge of BC, angle bad = angle ABC, angle ACD = angle ADC. Angle BAC = 78 degrees, angle DAC degree
∵ ADC is the external angle of abd
∴∠ADC＝∠ABC+∠BAD
∵∠BAD＝∠ABC
∴∠ADC＝2∠BAD
∵∠ADC＝∠ACD
∴∠DAC＝180-∠ACD-∠ADC＝180-2∠ADC＝180-4∠BAD
∵∠BAC＝∠BAD+∠DAC,∠BAC＝78
∴∠BAD+∠DAC＝78
∴∠DAC＝78-∠BAD
∴180-4∠BAD＝78-∠BAD
∴∠BAD＝34
∴∠DAC＝78-34＝44°
As shown in the figure, D is on the BC side of △ ABC, and ∠ abd = ∠ bad, ∠ ADC = ∠ ACD, ∠ BAC = 60 ° to find the degree of ∠ DAC
Let's set the equal angle as X, Y. let's solve the first-order equations of two variables
As shown in the figure, in △ BAC, D is a point on the edge of BC, angle bad = angle abd, angle ADC = angle ACD, angle BAC = 63 degrees, calculate angle DAC
Let the degree of the angle DAC be X,
Because the sum of the angles inside the triangle is 180 degrees,
The angle DCA is equal to (180-x) / 2 and the angle abd is equal to 63-x
So (180-x) / 2 + 63-x + 63 = 180
So x = 24 degrees
In the triangle ABC, point D is a point on BC, angle bad = angle abd, angle ADC = angle ACD, angle BAC = 63 degrees, calculate the degree of angle DAC
Because angle ADC = angle abd + angle bad, angle bad = angle abd
So angle ADC = 2 angle ABC
Because angle ADC = angle ACD
So angle ACD = 2 angle ABC
Because the angle BAC is 63 degrees
So the angle ABC + angle ACD = 180-63 = 117 degrees
Because angle ACD = 2 angle ABC
So the angle ABC = 117 / 3 = 39 degrees
Because angle bad = angle abd
So the angle bad is 39 degrees
Because the angle BAC is 63 degrees
So the angle DAC = angle BAC - angle bad = 63-39 = 24 degrees
In triangle ABC, point D is a point on BC, angle BAD= angle ABC, angle ADC= angle ACD. If angle BAC=63 degrees, find angle DAC and angle ADC
∠ADC=78°
∠DAC=24°
As shown in the figure, in △ ABC, ∠ C = 90 °, ad bisects ∠ BAC, and ∠ B = 3 ∠ bad, calculate the degree of ∠ ADC
Let ∠ bad be x! This is easier to explain. Because ∠ B = 3 ∠ bad, so ∠ B = 3x, and because ad bisects ∠ BAC, so ∠ bad + ∠ B = 5x = 180 ° - ∠ C = 90 ° and x = 18 ° is obtained. Therefore ∠ ADC = ∠ bad + ∠ B = 72 °
As shown in the figure. AB = ad, ∠ ABC = ∠ ADC, verification: BC = DC
Proof: connect BD,
∵AB=AD，
∴∠ABD=∠ADB．
∵∠ABC=∠ADC，
∴∠CBD=∠CDB．
∴BC=DC．
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