Question

Prove parallelogram side theorem for segments

Answer 1

STATEMENTS:

1.ABCD

2.AB cong. CD

3.BC cong. DA

4.DRAW AC

5.BCA and DAC are alt. angles

6.DCA and BAC are alt. angles

7.BAC cong. DCA

8. BCA cong. DAC

9. ABC cong. CDA

10.AB cong. CD

11. BC cong. DA

REASONS:

1. given

2. def. of parallelogram

3. def. of parallelogram

4. unique line postulate

5. def. of interior angles

6. def. of interior angles

7. alternate angles theorem

8. alternate angles theorem

9. reflexive property

10. ASA

11. CPCTC

* GLAD I COULD HELP HAVE A GOOD DAY !

Answer 2

The figure is attached.

To prove: In the ||gm ABCD, segment AB ≅ CD & segment BC ≅ AD.

Proof:

∵ ABCD is a parallelogram,
→ segment AB || segment DC
→ segment BC || segment AD
Now, the line AC is a transversal.
In Δ ABC and ΔADC∠1 = ∠4 (alternate interior angles)
similarly, ∠2 = ∠3 (alternate interior angles)
AC is common.
Hence, by ASA congruency property, we can say that,
Δ ABC ≅ ΔADC
⇒ segment AB ≅ CD & segment BC ≅ AD (congruent parts of congruent triangles)
Proved.