Question

Prove that a diagonal of a parallelogram divides it into two congruent triangles.​

Answer 1

Given: A parallelogram ABCD and AC is its diagonal .

To prove : △ABC ≅ △CDA

Proof : In △ABC and △CDA, we have

∠DAC = ∠BCA [alt. int. angles, since AD | | BC]

AC = AC [common side]

and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC]

∴ By ASA congruence axiom, we have

△ABC ≅ △CDA

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Answer 2

$\huge\colorbox{pink}{Question࿐ }$

Prove that a diagonal of a parallelogram divides it into two congruent triangles.

$\huge \orange{ \underline \red{ \underline \purple{Solution:-}}}$

GIVEN:-

• AB is a parallelogram.
• AC is a diagonal.

To prove:-

∆ ABC congruent ∆CDA

Proof: In ∆ ABC & ∆ CDA

• AB = BC [ Alternate Interior Angles ]
• BC = AD [ Alternate Interior Angles ]
• AC = AC ( Common )

∆ ABC congruent ∆ CDA ( by A.S.A [ Angle Side Angle ] ).

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