10 points! i need filling this in for school asap!

Given : EFGH is a parallelogram.

Prove: EG bisects HF and HF bisects EG.

Since, a parallelogram has two pairs of congruent and parallel sides.

Therefore, In parallelogram EFGH,

EF≅GH and EH≅FG

Also, EF║GH and EH║FG

Since,In triangles EKF and GKH,

EF≅HG

Since,  EF║HG

⇒∠FEK ≅ ∠HGK and ∠EFK ≅ ∠ GHK ( When two parallel lines are cut by a transversal alternative interior angles are congruent )

⇒ Δ EKF ≅ Δ GKH  ( ASA congruence postulate )

⇒ EK ≅ GK  and FK ≅ HK ( CPCTC)

⇒ EG bisects HF and HF bisects EG ( Definition of bisector)

Hence proved.

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

happy to

pls mark as brainliest