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3 years ago

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In parallelogram ABCD, one way to prove that diagonals and bisect each other is to prove triangle AXB congruent to triangle CXD.

Which triangle congruency theorem proves that the triangles are congruent?
ASA
SSS
SAS
AAS

2 answers:

DIA [1.3K]3 years ago

6 0

The answer is: AAS

Proving that two triangles are congruent by AAS can be verified when two angles and a non-included side in the first triangle are equal to their corresponding two angles and non-included side in the second triangle.

For the given parallelogram (check attachment):
angle AXB = angle CXD (opposite angle)
angle ABX = angle CDX (alternate angles)
AB = CD (opposite sides in parallelogram are equal)
Therefore,
<span>Triangle AXB is congruent to triangle CXD.</span>

GuDViN [60]3 years ago

3 0

Hello,

The answer is AAS

Hope this helps,

Lies‡

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raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

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Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

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We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

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$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

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${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

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We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

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The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

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