Prove this theorem by SSS rule.
"a diagonal of a parallelogram divides it into two congruent triangles"
By SSS rule only.
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Prove this theorem by SSS rule.
"a diagonal of a parallelogram divides it into two congruent triangles"
By SSS rule only.
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Given : A parallelogram ABCD.
To prove : ΔBAC ≅ ΔDCA
Construction : Draw a diagonal AC.
Proof :
In ΔBAC and ΔDCA,
∠1 = ∠2 [alternate interior angles]
∠3 = ∠4 [alternate interior angles]
AC = AC [common]
ΔBAC ≅ ΔDCA [ASA]
Hence, it is proved.