Prove equal chord of a circle substend equal angle at centre
1 answer:
Answer:
The Proof for "Equal chord of a circle subtends equal angle at the Center".
Step-by-step explanation:
Consider a Circle with center as C,
Radius AC ,BC ,DC and EC
Chord AB = Chord CD
To Prove:
"Equal chord of a circle subtends equal angle at center".
i.e ∠ACB = ∠ECD Which are Center angle
Proof:
In Δ ABC and Δ DEC
Chord AB ≅ Chord DE .….{Given}
AC ≅ DC …..{Radius of Same Circle}
BC ≅ EC ….{Radius of Same Circle}
Δ ABC ≅ Δ DEC ….{By Side-Side-Side Congruence Postulate}
∴∠ACB ≅ ∠ECD ...{Corresponding Parts of Congruent Triangles are Congruent or CPCTC} ....Proved
i.e Equal chord of a circle subtends equal angle at the Center.
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Answer: A sequence of similar transformations of dilation and translation could map △ABC onto △A'B'C'.
Step-by-step explanation:
Similar transformations: If one figure can be mapped onto the other figure using a dilation and a congruent rigid transformation or a rigid transformation followed by dilation then the two figures are said to be similar.
In the attachment △ABC mapped onto △A'B'C' by a sequence of dilation from origin and scalar factor k followed by translation.
