Proof Through Geometry
The figure shows an equilateral triangle ABC with DE parallel to BC and EF parallel to AB. D and G are on AB; F and H are on BC. GM and HN are perpendicular to AC. If AC = a, DG = d, FH = f, and MN = x prove that x = (a + d + f) / 2
Given:
Equilateral Triangle ABC
AC = a
DE // BC, EF // AB
DG = d
FH = f
GM perpendicular to AC
HN perpendicular to AC
Prove:
x=(a+d+f)/2
Given:
Equilateral Triangle ABC
AC = a
DE // BC, EF // AB
DG = d
FH = f
GM perpendicular to AC
HN perpendicular to AC
Prove:
x=(a+d+f)/2
Hint 1: DE // BC, EF // AB
By having parallel sides with our large equilateral triangle, ABC, we notice that triangle ADE and triangle EFC are also equilateral triangles. Therefore AD = DE = AE and
EF = FC = EC.
EF = FC = EC.
Hint 2: Other Than a Triangle
Notice the parallelogram and use it to find more equal lines.
Therefore AE = DE = BF and EC = EF = BD.
Therefore AE = DE = BF and EC = EF = BD.
Hint 3: Hint 3: Start with a
in a+d+f
Substitute a in for the line segments it represents from the triangle.
a = AE + EC
a = AE + EC
Hint 4: Create Midpoint k on a
There is now a line BK that splits the equilateral triangle in half.
We now have two 30-60-90 triangles. With these triangles come
special properties such as the hypotenus is twice as much as the base.
Therefore AB = 2AK and BC = 2KC
We now have two 30-60-90 triangles. With these triangles come
special properties such as the hypotenus is twice as much as the base.
Therefore AB = 2AK and BC = 2KC
