As shown in the figure, in the isosceles trapezoid ABCD, ab ∥ CD, DC = ad = BC, and the diagonal AC is perpendicular to the waist BC, calculate the degree of each internal angle of the trapezoid

As shown in the figure, in the isosceles trapezoid ABCD, ab ∥ CD, DC = ad = BC, and the diagonal AC is perpendicular to the waist BC, calculate the degree of each internal angle of the trapezoid

As shown in the figure, the crossing point C is CE ‖ ad, and DC ‖ AE, the 〈 quadrilateral AECD is parallelogram, and DC = ad = BC, the 〈 quadrilateral AECD is rhombus, the 〈 AE = CE = BC, the 〉 EAC = ∠ ECA, ∠ CEB = ∠ B, ∫ B + ∠ cab = 90 °, that is, 3 ∠ CAE = 90 °, the 〉 CAE = 30 °, the ∠ B = 60 ° = ∠ DAB, ∫ d = DCB = 120 °

In trapezoidal ABCD, ad ∥ BC, ab = DC, AE ⊥ ad intersects BD at point E, ∠ BAE = ∠ BDA

Extend AE to BC at point F,

In isosceles trapezoid ABCD, ab ‖ DC, ad = BC, if ∠ ADC = 105 ° diagonal AC is perpendicular to BD, then Tan ∠ DAC=

Method: make a parallel line of DB through point C, intersect the extension line of AB and E, get: because AC is perpendicular to BD, so AC is perpendicular to CE, because ab | CD, DB | Ce: DB = CE, so AC = CE, so the angle CAD is 45 degrees, because the angle ADC is 105 degrees, so the angle DAC is 30 degrees, and the tangent value is three thirds root

Be, CD are the heights of triangle ABC, proving that triangle ABC is similar to triangle AED

Angle CDA = angle bea = 90 degrees
Angle CAD = angle BAE
therefore
Triangle Abe ∽ triangle ACD
So AE: ad = AB: AC
that is
AE:AB=AD:AC
Another angle ead = angle bac
therefore
Triangle ade ∽ triangle ACB

As shown in the figure, in the triangle ABC, CD is perpendicular to AB and be is perpendicular to AC and E. connect De to prove that the angle AED is equal to the angle ABC
Similar 9 times

Where is the picture

As shown in the figure: given that ab = 10, points c and D are on the line AB and AC = DB = 2; & nbsp; P is the moving point on the line CD, with AP and Pb as edges respectively, make equilateral △ AEP and equilateral △ PFB on the same side of the line AB to connect EF, and let the midpoint of EF be g; when point P moves from point C to point D, the length of the moving path of point G is ()
A. 5B. 4C. 3D. 0

As shown in the figure, the intersection points of AE and BF are respectively extended at h. ∵ {a = {FPB = 60 °, ∵ ah} PF, ∵ {B =} EPA = 60 °, ∵ BH} PE, ∵ quadrilateral epfh is parallelogram, ∵ EF and HP are equally divided each other. ∵ G is the midpoint of EF, ∵ G is just the midpoint of pH, that is, in the process of P movement, G is always the midpoint of pH

As shown in the figure, we know Pb ⊥ Ba, PC ⊥ Ca, and Pb = PC, D is a point on PA, and prove: BD = CD

It is proved that: ∵ Pb ⊥ Ba, PC ⊥ Ca, in RT △ PAB, RT △ PAC, ∵ Pb = PC, PA = PA, ≌ RT △ PAB ≌ RT △ PAC, ≌ APB = ≌ APC, and D is the upper point of PA, PD = PD, Pb = PC, ≌ PBD ≌ PCD, ≌ BD = CD

It is known that ab = CD, which indicates that ab + CD is the middle term of a 2 + C 2 and B 2 + D 2

∵ AB = CD, ∵ ad = BC, ∵ (AB + CD) 2 = a2b2 + 2abcd + c2d2, (A2 + C2) (B2 + D2) = a2b2 + a2d2 + b2c2 + c2d2 = a2b2 + 2abcd + c2d2, ∵ (AB + CD) 2 = (A2 + C2) (B2 + D2), ∵ AB + CD is the median of the ratio of A2 + C2 and B2 + D2

In the triangle ABC, A-B = 4, a + C = 2B, and the maximum angle is 120 degrees

a=b+4=c+8 b=a-4=c+4 c=a-8=b-4
Let a be a diagonal a, B diagonal B, C diagonal C
Extend C out of a and make a vertical line from B to C. the remaining angle of fixing a is 60 degrees
aa-(c+b/2)(c+b/2)=bb-(b/2)(b/2)
The solution is: B = 10, a = 14, C = 6

In the triangle ABC, A-B = 4, a + C = 2B, and the maximum angle is 120 degrees?

a+c=2b a-b=b-c
∴a>b>c
a-b=4 (1)
a+c=2b (2)
a²=b²+c²-2bccos120°=b²+c²+bc (3)
From (1), a = 4 + B (4)
(4) Substituting (2) gives C = B-4 (5)
(4) (5) substitute (3) to get
(4+b)²=b²+(b-4)²+b(b-4)
B = 0 (rounding off) B = 10
The length of three sides is 14 10 6