Mathematics, 11.10.2019 14:30 Deliriousg636

In parallelogram abcd, e is the midpoint of ab and f is the midpoint of dc . let g be the intersection of the diagonal db and the line segment ef . prove that g is the midpoint of ef.

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Answer from: mlstogsdill05

The midpoint of the line $\overline{EF}$ is the point that divides $\overline{EF}$ in two halves of the same length.

ΔDFG ≅ ΔBGE and $\overline{FG}$ ≅ $\overline{EG}$ by CPCTC, therefore, G is the midpoint of $\overline{EF}$

Reasons:

The given parameters are;

The midpoint of AB in parallelogram ABCD = E

The midpoint of DC = F

Point of intersection of EF and DB = Point G

Required:

To prove that point G is the midpoint of EF.

Solution:

Statement Reason

1. m∠BDC ≅ m∠ABD 1. Alternate angles theorem

2. m∠DGF ≅ m∠BGE 2.Vertical angles theorem

3. $\overline{DC}$ = $\overline {AB}$ 3. Opposite sides of a parallelogram ABCD

4. $\overline{CF}$ ≅ $\overline{DF}$ 4. Definition of midpoint of DC

5. $\overline{CF}$ = $\mathbf{\overline{DF}}$ 5. Definition of congruency

6. $\overline{CF}$ + $\overline{DF}$ = DC 6. Segment addition property

7. $\overline{CF}$ + $\overline{CF}$ = DC 7. Substitution property

8. 2· $\overline{CF}$ = DC 8. Addition

9. $\overline{CF}$ = 0.5· $\overline{DC}$ = $\overline{DF}$ 9. Division property

Similarly;

10. $\overline{AE}$ = 0.5· $\overline{AB}$ = $\overline{EB}$ 10. Division property

11. 0.5· $\overline{DC}$ = 0.5· $\overline{AB}$ 11. Multiplication property of equality

12. $\overline{AE}$ = $\overline{EB}$ 12. Substitution property

13. ΔDFG ≅ ΔBGE 13. Angle-Angle-Side rule of congruency

14. $\overline{FG}$ ≅ $\overline{EG}$ 14. CPCTC

15. $\overline{FG}$ = $\overline{EG}$ 15. Definition of congruency

16. Point G is the midpoint of $\overline{EF}$ 17. Definition of midpoint

Learn more about the midpoint of a line here:

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Answer from: fri90

GF = GE that prove G is the mid-point of EF

Step-by-step explanation:

In the Parallelogram ABCD

∵E is the mid-point of AB

∵F is the mid-point of CD

∵AB = CD opposite sides in the parallelogram

∴EB = DF⇒(1)

∵AB // CD opposite sides in the parallelogram

∴m∠EBD = m∠FDB alternate angles ⇒(2)

∵BD intersects EF at G

∴m∠BGE = m∠DGF vertically opposite angles ⇒(3)

By using (1) , (2) and (3) you can prove:

ΔBGE is congruent to ΔDGF ⇒ AAS

∴GF = GE

∴G is the mid-point of EF

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In parallelogram abcd, e is the midpoint of ab and f is the midpoint of dc . let g be the intersecti...