Prove parallelogram side theorem for segments
1 answer:
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.
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.
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