Question

prove that the angle of a parallelogram divides it into two congruent triangles​

Answer 1

Answer:

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$\sf\green{prove \:that\: the \:angle \:of \:a \:parallelogram\: divides \:it\: into \: two \: congruent\: triangles}$

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☞Given:

A parallelogram ABCD.

☞To Prove:

A diagonal divides the parallelogram into two congruent triangles i.e., if diagonal AC is drawn then ∆ABC ≅ ∆CDA and if diagonal BD is drawn then ∆ABD ≅ ∆CDB

☞Construction:

Join A and C

☞Proof:

Sine, ABCD is a parallelogram AB ║ DC and AD ║ BC

In ∆ABC and ∆CDA

∠BAC = ∠DCA [Alternate angles]

∠BCA = ∠DAC [Alternate angles]

And, AC = AC [Common side]

$\boxed{∴ \: ∆ABC\: ≅\: ∆CDA\: [By\: ASA]}$

Similarly, we can prove that ∆ABD ≅ ∆CDB

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