Solution:
∫△ABC is an equilateral triangle and BD is the center line.
∴BD is divided equally ∠ABC (three lines are combined into one)
∴∠CBD= 1/2∠ABC=30
CE = CD
∴∠CED=∠CDE
∠∠CED+∠CDE =∠ACB (outer corner) = 60.
∴∠CED=∠CDE= 1/2∠ACB=30
∴∠CBD=∠CED=30
∴BD=DE
2. Fold a rectangular piece of paper ABCD along EF, with points D and C at D' and C' respectively, and the intersection of ED' and BC is G. If ∠ EFG = 55, find ∠ 1, ∠ 2. (It is estimated that this question is incomplete. If the data are different, ∞.
Solution:
As shown in the figure:
∵ quadrilateral ABCD is a rectangle
∴AD‖BC
∠∠EFG = 55
∴∠def =∞∠EFG = 55 (internal dislocation angles are equal)
∠ 1+∠ 2 = 180 (complementary to ipsilateral internal angle)
According to the nature of folding:
∠GEF=∠DEF=55
∴∠ 1= 180-∠ GEF -∠DEF=70
∴∠2= 180 -∠ 1= 1 10
3. In the isosceles triangle AB=AC, AB = AC, AD is the height on the side of BC, E and F are the points on the sides of AB and AC respectively, and EF‖BC.
(1) indicates that the triangle AEF is an equilateral triangle.
(2) Determine the shape of triangle DEF and explain the reasons.
Solution:
( 1)
∫△AB=AC is an isosceles triangle, and AB = AC.
∴∠ABC=∠ACB
∫EF‖BC
∴∠ AEF =∠ ABC, ∠AFE =∠ACB (same angle)
∴∠AEF=∠AFE
∴AE=AF
∴△AEF is an isosceles triangle
(2)
△DEF is an isosceles triangle for the following reasons:
∫△ABC is an isosceles triangle, and AD is the height on the side of BC.
∴AD bisection ∠BAC (three in one)
∴∠BAD=∠CAD
From (1), we can know: AE = AF.
Ad = ad again.
∴△AED≌△AFD(SAS)
∴ED=FD
△ def is an isosceles triangle.