В треугольнике АВС через точку пересечения биссектрис углов, прилежащих к основанию АС, проведена прямая, параллельная основанию и пересекающая боковые стороны в точках К и N. Найдите периметр треугольника KBN, если AB=20, BC=25, AC=18.
The perimeter of triangle KBN is 61.3 units.
To find the perimeter, we will first need to find the lengths of the sides KB and BN.
By drawing a diagram and labeling the angles and sides, we can see that triangle KBV and CBV share the same base, and angles KBV and CBV are vertical angles, making them equal.
Similarly, angles CNA and CKA are equal, since they are vertical angles sharing the same base in triangles ACN and ACK.
Therefore, we can conclude that triangle KBV and triangle ACK are similar, since they have two angles that are equal.
Using this similarity, we can set up a proportion between the sides of the triangles:
KB/AC = BN/BC
Solving for KB, we get: KB = (BN * AC)/BC
Substituting the given values, we get KB = (18 * 25) /18 = 25 units.
Similarly, we can find BN by setting up a proportion between triangles BNV and ABC and solving for BN.
Using the Pythagorean theorem, we can also find the length of VN, which is 24 units.
Now, we have all the required lengths to find the perimeter of triangle KBN, which is:
Perimeter(KBN) = KB + BN + KN = 25 + 24 + 12.3 = 61.3 units
