Question

Prove that in parallelogram opposite side are equal

Answer 1

Step-by-step explanation:

As an example, this proof has been set out in full, with the congruence test fully developed. Most of the remaining proofs however, are presented as exercises, with an abbreviated version given as an answer.

Theorem

The opposite sides of a parallelogram are equal.

Proof

ABCD is a parallelogram.

To prove that AB = CD and AD = BC.

Join the diagonal AC.

In the triangles ABC and CDA:

angleBAC = angleDCA (alternate

angles, AB || DC)

angleBCA = angleDAC

(alternate angles, AD || BC)

AC = CA (common)

so triangleABC ≡ triangleCDA (AAS)

Hence AB = CD and BC = AD

(matching sides of congruent triangles)

Answer 2

SOLUTION:

GIVEN: A parallelogram ABCD.

TO PROVE: A diagonal, say, AC, of parallelogram ABCD divides it into congruent triangles ABC and CDA i.e. ∠ABC ≅ ∠CDA.

CONSTRUCTION: Join AC.

PROOF: Since ABCD is a parallelogram. Therefore, AB || DC and AD || BC

Now, AD || BC and transversal AC intersects them at A and C respectively.

∴ ∠DAC = ∠BCA __[Alternate interior angles) ... (1)

Again, AB || DC and transversal AC intersects them at A and C respectively. Therefore,

∴ ∠BAC = ∠DCA __[Alternate interior angles) ...(2)

Now, in △s ABC and CDA, we have

∠BCA = ∠DAC ____ [From (1)]

AC = AC ________[Common side]

∠BAC = ∠DCA _____[From (2)]

So, by ASA congruence criterion, we obtain

△ABC ≅ △CDA ______[PROVED]

Also, AD = CB, DC = BA ___[ Corresponding parts of congruent triangles are equal ]