In parallelogram ABCD, the bisectors of angles B and C intersect side AD at points M and L, respectively. BC = 30, and lines BM and CL intersect at point P so that CP: CL = 5: 3. Find the AD side.
Since ABCD is a parallelogram, then BC is parallel to AD, and the angle BCP = MLP as the corresponding angles at the intersection of parallel BC and AD secant CP.
Then the triangles BCP and MLP are similar in two angles.
Let CP = 5 * X cm, then CL = 3 * X, and PL = 5 * X – 3 * X = 2 * X.
The coefficient of similarity of triangles is: K = 2 * X / 5 * X = 2/5.
Then ML / BC = 2/5.
ML = 2 * BC / 5 = 2 * 30/5 = 12 cm.
Since BP and CP are bisectors of the angles, they cut off the isosceles triangles ABM and CDL.
AB = CD = AM = DL.
Then AD = BC = 30 = 2 * AM + ML = 2 * AM + 12.
2 * AM = 30 – 12 = 18 cm.
AM = AB = 18/2 = 9 m.
Answer: The length of the AB side is 9 cm.
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