Solution: let c be the extension line from CE⊥BA to BA in e,
∫△ABC external angle is 80 degrees, and AB=AC.
∴∠BAC= 100
And CD equally ∠ACB
∴∠ADC=60
In Rt△DCE, CE= (root number 3)/2 times CD.
∴CE=(20/3)* root number 3× (root number 3)/2
= 10
That is waist circumference 10.
2. In a square ABCD, AD=8, point E is the moving point of the CD (excluding the end point), the middle vertical line FE of AE intersects with AD, AE and BC in F, H and K respectively, and the intersection point AB extends to point G. 。
1, let DE=m, FH/HK=t, and express t with an algebraic expression containing m.
2. When t= 1/3, find the length of BG.
Solution:
1. Crossing H means MN is parallel to AB, crossing AD is in M, and crossing BC is in N..
H is the midpoint of ∴ hm =1/2de =1/2m.ae.
And it is easy to prove △HMF∽△HNK.
∴MH/HN=FH/HK
That is, (1/2 * m)/(8-1/2 * m) = t.
∴t=m/( 16-m)
3. Triangle ABC is an equilateral triangle, D is any point on the side of BC, CE is the bisector of the outer corner of angle ACB, and angle ADE is equal to 60 degrees. Prove that AD=DE.
Certificate: DF⊥AC in F, DG⊥EC in G.
∠∠DCA =∠DCG = 60
∴DF=DG (properties of angular bisector)
It is also easy to prove ∠DAF =∠ degree.
∴△ADF≌△EDG(AAS)
∴AD=DE
4. Given a square ABCD, let the parallel line be AC passing through point B, let AE = AC, and let AE intersect BC at point F,
Verification: CE=CF
Certificate: connect BD to AC in O, cross E to EH⊥AC in H,
∫BE‖AC，
∴EH=BO= 1/2BD
And BD=AC and AE=AC.
∴EH= 1/2AE
∴∠eah = 30° (in a right triangle, if an acute angle equals 30, the right side it faces is equal to half of the hypotenuse).
AE=AC，∴∠ ACE = ∠ AEC = 75。
∠ ACB = 45。
∴∠ECF=75 -45 =30
∠CFE = 180-∠ECF-∠ACE = 180-75-30 = 75。
∴∠CFE=∠AEC
∴CE=CF
Please draw a picture first: a trapezoid, with the letter A above, D below B, and C in the order from left to right.
E is the midpoint of AD. Note: AD is waist. Don't draw the wrong picture.
Topic: (1) Angle A=90 degrees (2)AB+CD=BE (3) Area of triangle BEC = 1/2 Area of trapezoid ABCD (4) bisecting angle ABC (5) Angle BEC=90 degrees.
Please choose two related theories from the above five theories, take one as the condition and the other as the conclusion, construct a correct proposition and prove it.
It is known that in trapezoidal ABCD, e is the midpoint of AD and the angle A=90 degrees.
It is proved that the area of triangular BEC = 1/2 the area of trapezoidal ABCD.
Certificate: extend the extension line from BE to CD to F.
E is the midpoint of AD, ∴AE=DE,
And ∠ a = ∠ CDA = 90, ∠AEB=∠DEF.
∴△ABE≌△DFE
∴AB=FD
∴ area of trapezoid ABCD = (AB+DC) × AD/2 = (FD+DC) × AD/2 = area of △ BFC.
And S△EFC=[(FD+DC)×AD/2]/2=S△BFC/2.
∴S△EFC=S△BEC, that is, S△BEC=△BFC area /2= trapezoidal ABCD area /2.
6. In trapezoidal ABCD, AD is parallel to BC, AD=3, AB=4, BC=5, so what is the range of waist CD?
7. The height of the isosceles trapezoid is 6 cm and the diagonals are perpendicular to each other. What is the area of this trapezoid?
8 .. In the isosceles trapezoid ABCD, if AD is parallel to BC, the angle B=60 degrees, and AD=AB=6CM, what is the circumference of the isosceles trapezoid ABCD?
Solution:
6 .. If D is the intersection of DE‖AB and BC in E and ABED is a parallelogram, then CE=CB-BE=5-3=2.
DE=4,∴2