In the parallelogram ABCD, the point E is the midpoint of the side BC. The segment AE intersects the diagonal BD at the point K. a) Prove the similarity of the triangles AKD and EKB. b) Find the length of the segment AE if Ak is 7cm and the side BC is 16cm.
In triangles AKD and EKB, the angle AKB = EKB as the vertical angles at the intersection of straight lines AE and BD. Angle BEK = KAD as criss-crossing angles at the intersection of parallel straight lines BC and AD secant AE, then the triangles AKD and EKB are similar in two angles, which was required to be proved.
Since the current E is the middle of the BC, then BE = BC / 2 = 16/2 = 8 cm.
AD = 16 cm, since the opposite sides of the parallelogram are equal.
Then BE / AD = EK / AK.
8/16 = EK / 7.
EK = 8 * 7/16 = 3.5 cm.
AE = AK + EK = 7 + 3.5 = 10.5 cm.
Answer: The length of the segment AE is 10.5 cm.
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