Question

In the above figure, seg AB is a diameter of a circle with centre P. C is any point on the circle.  seg CE ⊥ seg AB. Prove that CE is the geometric mean of AE and EB. Write the proof with the help of the following steps:
a. Draw ray CE. It intersects the circle at D.
b. Show that CE = ED.
c. Write the result using the theorem of the intersection of chords inside a circle. d. Using CE = ED, complete the proof. 

Answer 1

In circle with centre P, AB is a diameter and seg CE ⊥ seg AB     ….(Given)

∴ seg PE ⊥ chord CD (Construction)

∴ sec CE ≅ seg DE     ….(i)

(Perpendicular drawn from the center of the circle to the chord bisects the chord.)

seg AB and seg CD are two chords intersecting inside the circle a point E

∴ AE × BE = CE × DE

(Theorem of internal division of chords)

AE × BE = CE × CE      .... (From i)

AE × BE = CE²

∴ CE² = AE × BE

Hence, it is proved that CE is the geometric mean of AE and EB.