Question

ii) In the figure seg AB is a diameter of the circle with centre P.
C is any point on the circle and Seg. CE I seg AB.
Prove that CE is the geometric of AE and
EB with help of the the following steps:
i) Draw ray CE to intersect the circle at D.
ii) Show that CE = ED
iii) Using CE = ED complete the proof.
iv) Hence find CE if AE = 9 and BE = 16.
B​

Answer 1

CE is the geometric Mean of AE  & BE , CE = 12

Step-by-step explanation:

CE⊥ AB

=> DE ⊥ AB

P is center of circle

Then

PC = PD = Radius

=> CE² = PC² - OE²

    DE² = PD² - OE²

PC = PD

=> CE² =   DE²

=> CE = DE

By using thorm of intersection of chord

AE * BE = CE * DE

=> AE * BE = CE * CE

=> AE * BE = CE²

=> CE = √AE * BE

CE is the geometric Mean of AE  & BE

AE = 9   BE = 16

CE = √9 * 16

=> CE = √144

=> CE = 12

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