Solving Triangle Side Segment Lengths with the Angle Bisector Theorem
The image depicts a geometrical problem involving a triangle with vertices labeled as A, B, and C. The sides opposite these vertices are labeled with their respective lengths: side BC measures 29 units, side AC measures 21 units, and side AB measures 20 units. There is also an angle bisector drawn from vertex B to side AC, dividing it into two segments (not labeled in the picture).
To solve for the unknown length of the segments into which the angle bisector divides side AC, we can use the Angle Bisector Theorem, which states that the angle bisector of a triangle divides the opposite side into two segments that are proportional to the adjacent sides.
Let x be the length of the segment from point A to the point where the bisector intersects side AC, and let (21 - x) be the length of the segment from that point to point C. According to the Angle Bisector Theorem:
x / (21 - x) = AB / BC
Substituting the given lengths:
x / (21 - x) = 20 / 29
Cross-multiplying gives:
29x = 20(21 - x)
29x = 420 - 20x
29x + 20x = 420
49x = 420
x = 420 / 49
x = 8.57 (rounded to two decimal places)
So, one segment is approximately 8.57 units long. To find the length of the other segment, we subtract this from the total length of side AC:
21 - 8.57 = 12.43 (rounded to two decimal places)
Thus, the angle bisector divides side AC into two segments approximately 8.57 units and 12.43 units in length.
