The chords AB and CD of the circle intersect at point E. AE: EB = 1: 7, C = 16cm, DE is 2 cm less than CE. Find AE.

| September 23, 2021 | education

Given: a circle, where E ∈ AB, E ∈ CD, CD = 16 cm, DE = CE – 2 cm and AE: EB = 1: 7. It is necessary to find the length of AE.
Obviously, DE + CE = СD, that is, CE – 2 cm + CE = 16 cm or 2 * CE = 16 cm + 2 cm, whence CE = (18 cm): 2 = 9 cm.Therefore, DE = CE – 2 cm = 9 cm – 2 cm = 7 cm.
We will use the following property of intersecting chords. The products of the lengths of the segments into which each of the chords is divided are equal. This fact for our assignment can be formalized as follows: AE * EB = CE * DE. We have AE * EB = (9 cm) * (7 cm) or AE * EB = 63 cm2.
Since AE: EB = 1: 7, then using the main property of proportion we have: EB = 7 * AE. Let’s substitute this expression in the last equality of item 3. Then, we have: AE * 7 * AE = 63 cm2 or AE2 = 63 cm2: 7 = 9 cm2. Therefore, AE = 3 cm.
Answer: AE = 3 cm.



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