Question

show that a diagonal of parallelogram divides it into two congruent triangles

Answer 1

Given:

Parallelogram = ABCD

Diagonal = AC

To Prove:

Diagonal of a parallelogram divides it into two congruent triangles

Solution:

In ΔABC and ΔACD

AB || CD  and AC is a transversal

Thus,

∠ ACB = ∠ CAD.

AC = AC (common side)

∠CAB = ∠ ACD.

Thus, by ASA congruency  ΔABC ≅ ΔACD

The corresponding part of the congruent  triangle are congruent

Therefore,

AB = CD + AD = BC

Hence proved

Answer 2

Answer:

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Step-by-step explanation:

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