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3 years ago

8

The midpoint of AC is

2 answers:

Rina8888 [55]3 years ago

7 0

Answer:

Step-by-step explanation:

Given ABCD is a parallelogram

To prove ⇒ AC trisects BD, and BD bisects AC.

Proof ⇒ Since coordinates of points A, B, C and D have been given in the diagram. If we prove that midpoint of AC and BD are common then AC and BD will equally bisect each other.

Midpoint of AC = $(\frac{x+x'}{2},\frac{y+y'}{2})$

coordinates of A and C are (0,0) and (2a + 2b, 2c)

Now mid point $E=(\frac{2a+2b+0}{2},\frac{2c+0}{2})$

= [(a+b),c]

Now mid point of BD = $(\frac{2b+2a}{2},\frac{2c+0}{2})$

= [(b+a), c]

It proves that midpoints of AC and BD are common.

So AC trisects BD and BD bisects AC.

AnnZ [28]3 years ago

5 0

The midpoint of $AC$ is $E$ .

Hope this helps.

r3t40

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Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

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Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

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$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

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