Question

prove that a diagonal of a parallelogram divides it into two congruent triangle​

Answer 1

consider Δ ABC and Δ ACD

Since the line segments AB+CD are parallel

to each other and AC is a transversal

∠ ACB = ∠ CAD.

AC = AC (common side)

∠CAB = ∠ ACD.

Thus, by ASA criteria

ΔABC ≅ ΔACD

The corresponding part of the congruent

triangle are congruent

AB = CD + AD = BC

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Answer 2

To prove:-

• A diagonal of a parallelogram divides it into two congruent triangles.

Now let us take a parallelogram with ABCD with AC as diagonal .

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Now , we need to prove that ΔABC ≅ ΔADC .

We know :

→ Opposite sides of parallelogram are equal .

∴ AB || DC & AD || BC .

As AB || DC (AC is transversal) :

∠BAC = ∠DCA .....(1)

(Alternate angles)

As AD || BC ,(AC is transversal) :

∠DAC = ∠BCA.....(2)

(Alternate angles)

From ΔABC & ΔADC ,

→ BAC = DCA ...(from 1 )

⇒ AC = AC (Common)

from (2) :

→ ∠ DAC =∠BCA

By ASA congruency ,

∴ ΔABC ≅ ΔADC

Therefore the diagonal of a parallelogram divides it into two congruent triangles.

Hence proved !

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