In Delta ABC,the correspondences ABC leftrigh | vedclass

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Question

(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

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(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.

Similar Questions

In $\Delta ABC$,$AB = 3$,$BC = 6$,and $AC = 8$. In $\Delta PQR$,$PQ = 13.6$,$QR = 10.2$,and $PR = 5.1$. Which correspondence between $\Delta ABC$ and $\Delta PQR$ is a similarity?

If $\triangle ABC \sim \triangle QRP$,$\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle QRP)} = \frac{9}{4}$,$AB = 18 \, cm$ and $BC = 15 \, cm$,then $PR$ is equal to (in $cm$):

Given $\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XYZ$. If $\frac{PQ}{3} = \frac{XY}{5}$ and $QR = 9$,find the length of $YZ$.

In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $\frac{AP}{PB} = \frac{3}{4}$ and $AC = 17.5$,find $AQ$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$,$AB = 2x + 3$,$BC = x + 2$,and $AC = 3x - 1$. Find the value of $x$.

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