If two equal chords of a circle
intersect within the circle, prove that the line joining the point, of
intersection to the centre makes equal angles with the chords.
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Hi there!
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Let's see some related topics :
⚫ Circle : The collection of all the points, which are at a fixed distance from a fixed point in a plane, is called a circle.
⚫ Radius : A line joining the centre to the Circumference of the circle, is called radius of a circle.
⚫ Secant : A line intersecting a circle at any two points, is called secant.
⚫ Diameter : A chord passing through the point of the circle, is called diameter. It is the longest chord.
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_______________________
For solutions, Refer to the attached picture.
Regrets for handwriting _/\_
_______________________
Let's see some related topics :
⚫ Circle : The collection of all the points, which are at a fixed distance from a fixed point in a plane, is called a circle.
⚫ Radius : A line joining the centre to the Circumference of the circle, is called radius of a circle.
⚫ Secant : A line intersecting a circle at any two points, is called secant.
⚫ Diameter : A chord passing through the point of the circle, is called diameter. It is the longest chord.
_______________________
Thanks for the question !
☺️❤️☺️
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Hello mate =_=
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Solution:
Let's suppose that we have circle with centre O. There are two equal chords AB and CD intersecting at point E.
Construction: Draw OM⊥AB and ON⊥CD. Join OE.
We need to prove that ∠OEM=∠OEN
In ∆OME and ∆ONE, we have
∠OME=∠ONE (Each equal to 90°)
OE=OE (Common)
OM=ON (Equal chords are equidistant from the centre)
Therefore, by RHS congruence rule, we have ∆OME≅∆ONE
⇒∠OEM=∠OEN (Corresponding parts of congruent triangles are equal)
hope, this will help you.
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