erama1
contestada

3.
Reasons
Statements
MNOP is a parallelogram
N
PM || ON
M
Given:
Alternate Int. Zs Thm,
MNOP is a parallelogram
MN II PO
Alternate Int. Zs Thm,
Prove:
PM = ON
(For this proof, use only the
definition of a parallelogram;
don't use any properties)
HE

3 Reasons Statements MNOP is a parallelogram N PM ON M Given Alternate Int Zs Thm MNOP is a parallelogram MN II PO Alternate Int Zs Thm Prove PM ON For this pro class=

Respuesta :

Answer:

3. [tex]\overline{PM}\cong \overline{ON}[/tex] Distances between two parallel lines [tex]\overline{MN} \ and\ \overline{PO}[/tex]

4. [tex]\overline{AC}[/tex] = [tex]\overline{CE}[/tex]: Reason; Corresponding part of ΔACB and ΔDCE

C is the midpoint of [tex]\overline{AE}[/tex]: Reason;  [tex]\overline{AC}[/tex] = [tex]\overline{CE}[/tex]: Definition of midpoint

Step-by-step explanation:

3. A parallelogram is defined as a quadrilateral with two opposite sides equal and parallel and having equal opposite interior angles

MNOP is a parallelogram: Reason; Given

[tex]\overline{PM}\left | \right |\overline{ON}[/tex] : Reason; Opposite sides of a parallelogram

∠NOM ≅ ∠OMP: Reason Alternate interior angles

[tex]\overline{MN}\left | \right |\overline{PO}[/tex]: Reason; Opposite sides of a parallelogram

∠NMO ≅ ∠MOP: Reason Alternate interior angles

[tex]\overline{PM}\cong \overline{ON}[/tex] Distances between two parallel lines [tex]\overline{MN} \ and\ \overline{PO}[/tex]

4. [tex]\overline{AB}\left | \right |\overline{DE}[/tex] : Reason; Given

∠EAB ≅ ∠AED: Reason; Alternate int. ∠s Thm

∠ABC ≅ ∠EDB : Reason; Alternate int. ∠s Thm

C is the midpoint of  [tex]\overline{BD}[/tex]: Reason; Given

[tex]\overline{BC}[/tex] = [tex]\overline{CD}[/tex]: Reason; Definition of midpoint

Therefore, ΔACB ≅ ΔDCE: Reason Angle Angle Side (AAS) Theorem

[tex]\overline{AC}[/tex] = [tex]\overline{CE}[/tex]: Reason; Corresponding part of ΔACB and ΔDCE

C is the midpoint of [tex]\overline{AE}[/tex]: Reason; Definition of midpoint