In triangles ABC and PQR,AB = AC,angle C = an | vedclass

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(C) Given: In $	riangle ABC$,$AB = AC$. Therefore,$\angle B = \angle C$ (Angles opposite to equal sides are equal).Given: $\angle C = \angle P$ and $

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isosceles but not congruent

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(C) Given: In $	riangle ABC$,$AB = AC$. Therefore,$\angle B = \angle C$ (Angles opposite to equal sides are equal).Given: $\angle C = \angle P$ and $\angle B = \angle Q$.Since $\angle B = \angle C$,we have $\angle Q = \angle P$.In $	riangle PQR$,since $\angle Q = \angle P$,the sides opposite to these angles must be equal,so $PR = QR$. Thus,$	riangle PQR$ is an isosceles triangle.However,we do not have enough information (such as a side length or a specific angle measure) to prove that $	riangle ABC \cong 	riangle PQR$. For example,the triangles could have different sizes while maintaining the same angle relationships.Therefore,both triangles are isosceles,but they are not necessarily congruent.

In triangles $ABC$ and $PQR$,$AB = AC$,$\angle C = \angle P$ and $\angle B = \angle Q$. The two triangles are

Similar Questions

State whether each of the following statements is true or false:
$(1)$ In $\Delta XYZ$,if $XY > XZ$,then $\angle Z > \angle Y$.
$(2)$ In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PR} = \frac{BC}{QP} = \frac{CA}{RQ} = 1$,then $\Delta ABC \cong \Delta RPQ$.

Is it possible to construct a triangle with lengths of its sides as $9 \, cm$,$7 \, cm$,and $17 \, cm$? Give reason for your answer.

$AD$ is a median of the triangle $ABC$. Is it true that $AB + BC + CA > 2AD$? Give reason for your answer.

If $\Delta ABC \cong \Delta RPQ$ and the perimeter of $\Delta ABC$ is $18 \, cm$,then the perimeter of $\Delta PQR$ is $cm$.

$ABC$ is an isosceles triangle in which $AC = BC$. $AD$ and $BE$ are respectively two altitudes to sides $BC$ and $AC$. Prove that $AE = BD$.

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