prove that the diogonal of parallelogram divides it into similar triangles
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Step-by-step explanation:
REF. Image.
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:
Parallelogram Theorem #1: Each diagonal of a parallelogram divides the parallelogram into two congruent triangles. Parallelogram Theorem #2: The opposite sides of a parallelogram are congruent. ... You can show that alternate interior angles are congruent and hence lines are parallel for this part of the proof.
Step-by-step explanation:
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