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.
Answers
Answered by
32
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 !
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Answered by
1
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
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