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The answer is "No Solution" because there is not enough information.
the midpoint of
Yes. AB, AC, BC and EF.
The problem is meaningless without a diagram but I am guessing that ABC make a triangle and D is on the extension of AB beyond B. In that case we use the exterior angle theorem to get CBD = C + A, so 125 = 90 + A and A = 35.
ac + cb = ab = 9 2x - 1 + 3x = 9 5x -1 = 9 So 5x = 10 Thereby x =2. Also ac = 3 and cb = 6
14
If 2 segments have the same length they are known as 'congruent segments' IE: segment AB=segment AC (or AB=AC) then AB @ AC (or AB is congruent to AC)
segment ac
The answer is "No Solution" because there is not enough information.
Ab+bc=ac
Since B is located between A and C, you can just add the two lengths together, so AC = m + n.your segment looks like this:A----B----Cwhere AB=m, BC=n, and AC=m+n
Usually a line segment in a triangle is either assigned a letter, or is referred to by the letters at the end of the segment with a line overhead. For instance, if you have a triangle ABC, you'll have segments AB, AC, and BC. And there would be a line over each segment name.
It is easiest to draw it using two right angled triangles.Draw a line AB that is 2 units long. From B, draw BC which is perpendicular to AB and 2 units long. Join AC. From C, draw CD which is perpendicular to AC (clockwise if BC is clockwise from AB, or anticlockwise if BC is anticlockwise) and make CD 2 uinits long. Then AD is a line segment which is sqrt(12) units long.
Assuming A, B and C are consecutive points on a straight line, AB would be 49 mm
C is not on the line AB.
if segment ab is congruent to segment CD then segment ac is congruent to segment bd (only if points a, b, c, and d are all collinear)
C is not on the line AB.