
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Answer
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Hint: When two triangles are congruent they will have exactly the same three sides and exactly the same three angles. The equal sides and angles may not be in the same position (if there is a turn or a flip), but they are there.
SSS stands for "side, side, side" and means that we have two triangles with all three sides equal.
If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent.
Complete step-by-step answer:
To Prove: \[\Delta {\text{ }}ABC \sim \Delta DEF\]
Draw PQ in triangle DEF so that, AB = DP and AC = DF
Proof:
As all the three sides of two triangles are equal and one angle corresponding to the sides are equal. So, both the triangles are congruent.
\[\Delta ABC \cong \Delta DPQ\]
Because corresponding sides of these two triangles are equal
$\dfrac{{AB}}{{DE}} = \dfrac{{AC}}{{DF}}$ {Already given in the question}
$\angle A = \angle D${Already given in the question}
Hence; $\dfrac{{AB}}{{DE}} = \dfrac{{BC}}{{EF}}$ from SSS criterion
Hence;
$\dfrac{{AB}}{{DE}} = \dfrac{{AC}}{{DF}} = \dfrac{{BC}}{{EF}}$
Hence; \[\Delta {\text{ }}ABC \sim \Delta DEF\] proved.
Note: SAS (Side-Angle-Side)
If two pairs of corresponding sides are in proportion, and the included angle of each pair is equal, then the two triangles they form are similar. Any time two sides of a triangle and their included angle are fixed, then all three vertices of that triangle are fixed. With all three vertices fixed and two of the pairs of sides proportional, the third pair of sides must also be proportional.
SSS stands for "side, side, side" and means that we have two triangles with all three sides equal.
If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent.
Complete step-by-step answer:
To Prove: \[\Delta {\text{ }}ABC \sim \Delta DEF\]
Draw PQ in triangle DEF so that, AB = DP and AC = DF
Proof:
As all the three sides of two triangles are equal and one angle corresponding to the sides are equal. So, both the triangles are congruent.
\[\Delta ABC \cong \Delta DPQ\]
Because corresponding sides of these two triangles are equal
$\dfrac{{AB}}{{DE}} = \dfrac{{AC}}{{DF}}$ {Already given in the question}
$\angle A = \angle D${Already given in the question}
Hence; $\dfrac{{AB}}{{DE}} = \dfrac{{BC}}{{EF}}$ from SSS criterion
Hence;
$\dfrac{{AB}}{{DE}} = \dfrac{{AC}}{{DF}} = \dfrac{{BC}}{{EF}}$
Hence; \[\Delta {\text{ }}ABC \sim \Delta DEF\] proved.
Note: SAS (Side-Angle-Side)
If two pairs of corresponding sides are in proportion, and the included angle of each pair is equal, then the two triangles they form are similar. Any time two sides of a triangle and their included angle are fixed, then all three vertices of that triangle are fixed. With all three vertices fixed and two of the pairs of sides proportional, the third pair of sides must also be proportional.
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