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.
triangle ADE and triangle EFC are also equilateral triangles. Therefore AD = DE = AE
and EF = FC = EC.
