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Sedbober [7]
3 years ago
12

Prove parallelogram side theorem for segments

Mathematics
1 answer:
spin [16.1K]3 years ago
6 0
The figure is attached.

To prove: In the ||gm ABCD, segment AB ≅ CD & segment BC ≅ AD.

Proof:

∵ ABCD is a parallelogram,
→ segment AB || segment DC 
→ segment BC || segment AD 
Now, the line AC is a transversal.
In  Δ ABC and ΔADC∠1 = ∠4 (alternate interior angles)
similarly, ∠2 = ∠3 (alternate interior angles)
AC is common.
Hence, by ASA congruency property, we can say that, 
Δ ABC ≅ ΔADC
⇒ segment AB ≅ CD & segment BC ≅ AD (congruent parts of congruent triangles)
Proved.

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Answer: Yes, it's true.

Step-by-step explanation:

The first step is to convert the mixed number 1\frac{2}{5} to a fraction. The procedure for this is:

  1. Multiply the whole number 1 by the denominator 5.
  2. Add the product obtained and the numerator 2 (This will be the numerator of the fraction).
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Then:

1\frac{2}{5}=\frac{(1)(5)+2}{5}=\frac{7}{5}

Rewrite the equation:

\frac{7}{5}-\frac{3}{4}=\frac{1}{4}+\frac{2}{5}

Notice that the denominators on the left side are 4 and 5 and the denominator on the right side are also 4 and 5.

Then, the Least Common Denominator (LCD) on both sides is:

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LCD=5*2^2=20

Now we can solve the subtraction on the left side of the equation and the addition on the righ side:

\frac{(7*4)-(3*5)}{20}=\frac{(1*5)+(2*4)}{20}\\\\\frac{13}{20}=\frac{13}{20}

Therefore:

1\frac{2}{5}-\frac{3}{4}=\frac{1}{4}+\frac{2}{5}  (IT'S TRUE)

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