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In triangle ABC, the bisector of B and C angle meet P. Through P a straight line MN is drawn parallel to BC. Prove that MN=BM+CN

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Assuming M is the intersection of MN with AB, and N is the intersection of MN and AC:

Angle ACP = angle BCP (by definition)
Angle NCP = angle BCP (intersection of line with parallel lines produces equal angles)
Triangle CPN is isoceles (two equal angles), and line NP = CN

Same argument for line MP = BM

Therefore NP + MP (i.e, MN) = CN + BM

Posted on Sep 08, 2009

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If M is the midpoint of triangle ABC, ACB = 30degrees and BMA = 45 degrees, calculate the size of BAC in degrees.


Since M is the mid point of the triangle, the distance of M from A,B and C is equal which means AM = BM = CM.
Therefore, MAB = MBA. In triangle MAB, MAB=45 degrees.
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We can use BC or AB as the base.

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We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

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We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

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If you want the mathematical description of the sine of an angle it is described as follows
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