In Delta ABC,the correspondences ABC leftrigh | vedclass

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(A) Given that the correspondences $ABC \leftrightarrow BAC$ and $ABC \leftrightarrow ACB$ are similarities.From $ABC \leftrightarrow BAC$,we have $\a

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equilateral

Vedclass · Answer

(A) Given that the correspondences $ABC \leftrightarrow BAC$ and $ABC \leftrightarrow ACB$ are similarities.From $ABC \leftrightarrow BAC$,we have $\angle A = \angle B$,$\angle B = \angle A$,and $\angle C = \angle C$.From $ABC \leftrightarrow ACB$,we have $\angle A = \angle A$,$\angle B = \angle C$,and $\angle C = \angle B$.Combining these,we get $\angle A = \angle B$ and $\angle B = \angle C$.Therefore,$\angle A = \angle B = \angle C$.Since all three angles are equal,the triangle must be equiangular.An equiangular triangle is always an equilateral triangle.Thus,$\Delta ABC$ is an equilateral triangle.

In $\Delta ABC$,the correspondences $ABC \leftrightarrow BAC$ and $ABC \leftrightarrow ACB$ are similarities. Then,$\Delta ABC$ is a/an $\ldots \ldots \ldots \ldots$ triangle.

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