Question

In the adjoining figure, seg AB is a diameter of a circle with centre P. C is any point on
the circle. seg CE bisects seg AB. Prove that CE is the geometric mean of AE and EB. Write
the proof with the help of following steps:
(i) Draw ray CE. It intersects the circle at D.
(ii) Show that CE = ED.
(111) Write the result using theorem of intersection of chords inside
a circle.
(iv) Using CE = ED, complete the proof.
​

Answer 1

Answer:

Here is your proof!

Step-by-step explanation:

To prove :

CE2 = AE

EB

Construction : Draw ray CE, which intersects at point Don circle

Proof Now EP |_ CD, CE = ED

(1) chord, bisects it ] ..* perpendicular to 2

By using theorem of intersection of chords inside the circle,

AE x EB = CEX ED

AEX EB = CEXCE

From 1) (E= ED

. CE2 = AE EB

Hence, it is prove !

Answer 2

Answer:

it is the answer...

Step-by-step explanation:

To prove:

EP |_ CD, CE = ED

Construction: Draw ray CE, which intersects at point Don circle

Proof: Now EP |_ CD, CE = ED

(1) Chord, bisects it]...* perpendicular to 2

By using theorem of intersection of chords inside the circle,

AE x EB = CE X ED

AE X EB = CE X CE

From 1) (E= ED,

therefore, 2CE = AE EB

Hence, it is prove