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It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
In this paper, we have the idea that ᙽ EBD = x
As shown in the figure, in △ ABC, D is a point on the edge of BC, ad = BD, ab = AC = CD, and find the degree of ∠ BAC
∵AD=BD
ν let ∠ bad = ∠ DBA = x °,
∵AB=AC=CD
∴∠CAD=∠CDA=∠BAD+∠DBA=2x°，∠DBA=∠C=x°，
∴∠BAC=3∠DBA=3x°，
∵∠ABC+∠BAC+∠C=180°
∴5x=180°，
∴∠DBA=36°
∴∠BAC=3∠DBA=108°．
In the triangle ABC, D and E are the points on the edge of BC, BD = AB, CE = AC, and the angle DAE = 1 / 2, angle BAC. Find the degree of angle BAC ABC is a large triangle, in which there are two segments connected to point a and another point on line BC. These two segments are called AE and AD
Because BD = AB, CE = AC, so the angle bad = angle BDA, angle EAC = angle AEC, because angle ead + angle AED + angle ade = 180 degrees, angle EAC = angle ead + angle DAC, angle bad = angle ead + angle BAE, so angle ead + ead + ead + DAC = 180 degrees, because angle DAE = 1,2 angle BAC = 180 degrees, BAC = 90 degrees
As shown in the figure, in RT △ ABC, ∠ BAC = 90 °, AC = AB, ∠ DAE = 45 ° and BD = 3, CE = 4, find the length of de
As shown in the figure, rotate △ AEC clockwise around point a to △ AFB, and connect DF;
∵ △ ABC is an isosceles right triangle
∴∠ABD=∠C=45°；
And ∵ AFB ≌ △ AEC,
∴BF=EC=4，AF=AE，∠ABF=∠C=45°；
∵∠ABD=45°，
∴∠DBF=∠ABD+∠ABF=90°，
The △ DBF is a right triangle,
According to Pythagorean theorem, df2 = BF2 + BD2 = 42 + 32 = 52
∴DF=5；
Because ∠ DAE = 45 °, so ∠ DAF = ∠ DAB + ∠ EAC = 45 °;
∴△ADE≌△ADF（SAS）；
∴DE=DF=5．
As shown in the figure, the vertical bisectors of AB and AC on both sides of △ ABC intersect BC respectively at D and E. if ∠ BAC + ∠ DAE = 150 °, calculate the degree of ∠ BAC
∵ △ the vertical bisectors of AB and AC on both sides of ABC intersect BC at D, e, respectively,
∴DA=DB，EA=EC，
∴∠B=∠DAB，∠C=∠EAC．
∵∠BAC+∠DAE=150°，①
∴∠B+∠C+2∠DAE=150°．
∵∠B+∠C+∠BAC=180°，
∴180°-∠BAC+2∠DAE=150°，
That is ∠ bac-2 ∠ DAE = 30 °
The system of equations composed of ① and ②
∠BAC+∠DAE＝150°
∠BAC−2∠DAE＝30° ，
The solution is ∠ BAC = 110 °
So the answer is: 110 degrees
As shown in the figure, in the triangle ABC, ab = AC, D is one side of AC and ad = BD = BC, calculate the degree of ∠ ADB
∵AD=BD,∴∠A=∠ABD（1）
And BD = BC, C = ∠ BDC (2)
∴∠C=2∠A,
∠C=∠A+∠CBD,
∴∠A=∠CBD=1/2∠C,
From ∠ a + 2 ∠ C = 180 °,
5∠A=180°,
∴∠A=36°,
∴∠ADB=180-36°×2=108°.
As shown in the figure, in △ ABC, D is a point on the edge of BC, ad = BD, ab = AC = CD, and find the degree of ∠ BAC
∵AD=BD
ν let ∠ bad = ∠ DBA = x °,
∵AB=AC=CD
∴∠CAD=∠CDA=∠BAD+∠DBA=2x°，∠DBA=∠C=x°，
∴∠BAC=3∠DBA=3x°，
∵∠ABC+∠BAC+∠C=180°
∴5x=180°，
∴∠DBA=36°
∴∠BAC=3∠DBA=108°．
The center of circle O is the circumscribed circle of triangle ABC, and ab = AC, point d moves on arc BC, passing through point D as de / / BC, de intersects the extension line of AB at point E, connecting AD.BD . Then, when AB = 5. BC = 6, the radius of center O is calculated
This is a strange question,
Point d moves on arc BC, passing through point D as de / / BC, the extension line of de intersecting AB is connected with point E AD.BD .
None of these conditions worked
It can be found that the center of circle O is the circumcircle of triangle ABC and ab = AC, ab = 5. BC = 6
r=25/8
As shown in the figure, ⊙ o is the circumscribed circle of ⊙ ABC, and ab = AC, point d moves on arc BC, passing through point D as de ∥ BC, and the extension line of de intersecting AB at point e connects AD and BD (1) It is proved that: ∠ ADB = ∠ E; (2) When AB = 6, be = 3, find the length of AD
(1) It is proved that: ∵ AB = AC, point d moves on arc BC, passing through point D as de ∵ BC AB = AC, ∠ ABC = ∠ AED, ∠ ABC = ∠ ACB, ∠ ADB = ∠ e (2) ∵ ABC = ∠ AED, ∠ ABC = ∠ ACB, ∠ ADB = ∠ ACB,
In the triangle ABC, ab = 5, BC = 8, ∠ ABC = 60 °, D is a point on the AC arc of its circumscribed circle, and CD = 3, then the length of ad is The result is OK
AD=5
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