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Leokris [45]
3 years ago
10

Given: AB ≅ BC , m∠MOC = 135° OM − angle bisector of ∠AOB Prove: ∠ABO ≅ ∠CBO

Mathematics
1 answer:
Gennadij [26K]3 years ago
8 0

Answer:

See explanation

Step-by-step explanation:

1. Angles AOM and MOC are supplementry angles. If m∠MOC = 135°, then

m∠AOM = 180° - 135° = 45°

2. OM − angle bisector of ∠AOB, then

m∠AOM = m∠MOB = 45°

3. Now

m∠BOC = m∠MOC - m∠MOB

m∠BOC = 135° - 45° = 90°

4. Since m∠BOC = 90°, BO is perpendicular to AC.

5. Consider isosceles triangle ABC (because AB ≅ BC). BO is the height drawn to the base, so it is an angle B bisector too, thus

∠ABO ≅ ∠CBO

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On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.
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Answer:

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No, the distance from (0, 0) to (2, √6) is not 3 units

Step-by-step explanation:

The given parameters of the question are as follows;

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Point on the circle (x, y) = (3, 0)

We are required to verify whether point (2, √6) lie on the circle

We note that the radius of the circle is given by the equation of the circle as follows;

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We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;

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Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point  (2, √6), does not lie on the circle.

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