in the figure C (O,r) and C (O,r/2) touch internally at point a and ab is a chord of a circle so are intersecting C o, r/2at C prove that an AC is equal to cb
Answers
Answer:
1009682
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Medium
Solution
verified
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Join OA,OC and OB.
Clearly, ∠OCA is the angle in a semi-circle.
∴ ∠OCA=90
o
In right triangles OCA and OCB, we have
OA=OB=r
∠OCA=∠OCB=90
o
and, OC=OC
So, by RHS−criterion of congruence, we get
△OCA≅△OCB
∴AC=CB1009682
expand
Medium
Solution
verified
Verified by Toppr
Join OA,OC and OB.
Clearly, ∠OCA is the angle in a semi-circle.
∴ ∠OCA=90
o
In right triangles OCA and OCB, we have
OA=OB=r
∠OCA=∠OCB=90
o
and, OC=OC
So, by RHS−criterion of congruence, we get
△OCA≅△OCB
∴AC=CB
Step-by-step explanation:
1009682
expand
Medium
Solution
verified
Verified by Toppr
Join OA,OC and OB.
Clearly, ∠OCA is the angle in a semi-circle.
∴ ∠OCA=90
o
In right triangles OCA and OCB, we have
OA=OB=r
∠OCA=∠OCB=90
o
and, OC=OC
So, by RHS−criterion of congruence, we get
△OCA≅△OCB
∴AC=CB1009682
expand
Medium
Solution
verified
Verified by Toppr
Join OA,OC and OB.
Clearly, ∠OCA is the angle in a semi-circle.
∴ ∠OCA=90
o
In right triangles OCA and OCB, we have
OA=OB=r
∠OCA=∠OCB=90
o
and, OC=OC
So, by RHS−criterion of congruence, we get
△OCA≅△OCB
∴AC=CB