Given: BFCE, AB perpendicular to BE, DE perpendicular to BE, angle BFD equal to angle ECA
Prove triangle ABC is similar to triangle DEF

Step-by-step explanation:

From the given picture,

∠ABE = ∠DEF = 90° [Since, AB and DE are perpendicular to DE]

m∠ECA = m∠BFD [Given]

m∠ECA + m∠ACB = 180° [Liner pair of angles]

m∠BFD + m∠DFE = 180° [Liner pair of angles]

m∠ACB + m∠ECA = m∠BFD + m∠DFE [Transitive property]

m∠ACB = m∠DEF [Since, m∠ECA = m∠BFD]

Therefore, ΔABC ≅ ΔDEF [By AA property of similarity]

29 mil

step-by-step explanation:

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb