In the triangle ABC, the points D and E are on the sides AB and AC respectively, and the point F is on the extension line of De, and CF / / AB, ad * EF = BD * De

In the triangle ABC, the points D and E are on the sides AB and AC respectively, and the point F is on the extension line of De, and CF / / AB, ad * EF = BD * De

In △ AED and △ CFE
Because angle AED = angle CEF
AB parallel CF
Angle a = angle ECF
So △ AED ∽ CFE
So de: EF = ad: CF
From ad × EF = BD × de
AD:BD=DE：EF
And because de: EF = ad: CF
So ad: BD = de: EF = ad: CF
So ad: BD = ad: CF
So BD = CF
And because BD is parallel to CF
So a quadrilateral BCFD is a parallelogram
So de parallels BC
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In the triangle ABC, D and E are the midpoint of AB and AC respectively, f is a point on the extension line of BC, CF = 1 / 2BC, DC = EF

It is proved that: De is the median line of triangle
‖ de ‖ and = 1 / 2BC
And ∵ CF = 1 / 2BC
∴DE=CF
∵DE‖BC
The quadrilateral DCFE is a parallelogram
∴DC=EF

If x = 2, y = - 1 is a solution of binary linear equations ax + by = - 2, then 2a-b + 7

2a-b=-2
2a-b+7=5

Calculation: (1) 0.125 ^ 25 × 2 ^ 75 (2) 2 (x ^ 3) ^ 2 · x ^ 3 (3x ^ 3) ^ 3 + (5x)

（1）0.125^25×2^75
=0.125^25×8^25
=(0.125×8)^25
=1
（2）2（x^3)^2·x^3(3x^3)^3+(5x)
= 2x^6 ×x^3 ×27x^9+5x
=54x^18+5x