Question

Drag and drop a statement or reason to each box to complete the proof.

Answer 1

Answer:

Given : EFGH is a Parallelogram

Prove : EG Bisects HF , HF Bisects EG

Step-by-step explanation:

Proof

Check image below

Answer 2

Answer:

Given Parallelogram EFGH

EG bisects HF and HF bisects EG, if and only if both the diagnols have same mid point.

Step-by-step explanation:

Step 01:

Let

E be the point (a,b)

F be the point (a',b)

G be the point (a',b')

H be the point (a,b')

Step 02:

Now find mid points of EG and HF

mid point of EG = ( $\frac{a+a'}{2}$, $\frac{b+b'}{2}$ ) and

mid point of HF = (  $\frac{a'+a}{2}$,$\frac{b'+b}{2}$ )

Since addition is commutative, and they have the same mid-point, so they bisect each other.