Question

if each of the two parallel chords of a circle are bisected by a thrid chords then prove that the chords is a the diameter of the chord​

Answer 1

Answer:

Step-by-step explanation:

Let AB and CD be two chords intersecting at point O. Join AC and BD.

Now ΔAOC≈ΔBOD

⇒AC=BD

⇒∧AC =∧ BD-----------(1)

Now,

 ΔAOD≈ΔBOC

⇒AD=BC

⇒∧AD=∧BC--------(2)

(1)+(2)

⇔∧AC+∧AD =∧BD+∧BC

⇒ ANGLE "CAD"= ANGLE "CBD"

Then CD divides the circle into two equal parts thus CD is a diameter

Similarly AB is also the diameter and they both meet at point O

Thus they bisect each other....