A quadrilateral ABCD is inscribed in a circle such that AB:CD = 2:1 and BC:AD = 5:4. If AC and BD intersect at the point E, then AE:CE equals
Question:
A quadrilateral ABCD is inscribed in a circle such that AB:CD = 2:1 and BC:AD = 5:4. If AC and BD intersect at the point E, then AE:CE equals
A quadrilateral ABCD is inscribed in a circle such that AB:CD = 2:1 and BC:AD = 5:4. If AC and BD intersect at the point E, then AE:CE equals
Options
Answer: 8:5
Explanation:
In cyclic quadrilateral, by power of point theorem: AE/EC = (AB×AD)/(CB×CD). Given AB=2k, CD=k, BC=5m, AD=4m. Ratio = (2k×4m)/(5m×k) = 8m/5m = 8/5.
Explanation:
In cyclic quadrilateral, by power of point theorem: AE/EC = (AB×AD)/(CB×CD). Given AB=2k, CD=k, BC=5m, AD=4m. Ratio = (2k×4m)/(5m×k) = 8m/5m = 8/5.
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