Given: segment ac is perpendicular to segment bd prove: δacb ~ δecd reflect δecd over segment ac. this establishes that then, this establishes that ∠e'd'c' ≅ ∠abc. therefore, δacb ~ δecd by the aa similarity postulate. ∠abc ≅ ∠e'd'c'; translate point e' to point a ∠acb ≅ ∠e'c'd'; translate point e' to point b ∠acb ≅ ∠e'c'd'; translate point d' to point b ∠abc ≅ ∠e'd'c'; translate point d' to point a

easy math just plug in a and b

step-by-step explanation:

1e+27

step-by-step explanation:

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)