Congruence of Triangles By SSS Criteria
Congruence of Triangles by SSS Criteria
Theorem: Two triangles are congruence if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given: Two such that AB = DE, BC = EF and AC = DF.
To Prove:
Construction: Draw a line segment EG on the other side of EF such that AB = EG and . Join GF and GD.
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Proof: In BC = EF [Given] AB = GE [By construction] and, So, by SAS criterion of congruence, we have
Now, AB = DE and AB = GE
Similarly, AC = DF and AC = GF In DE = GE ....(iii) In DF = GF [From (ii)]
From (iii) and (iv), we have
But, Thus, in AB = DE [Given] and, AC = DF [Given] So, by SAS criterion of congruence , we have
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Illustration: ABCD is a parallelogram , if the two diagonals are equal, find the measure of
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Solution: Since ABCD is a parallelogram. Therefore, AB = CD and AD = BC [ Thus, in AD = BC [As proved above] BD = AC [Given] and, AB = AB [common] So, by SSS criterion of congruence, we have Now,
transversal is Hence, the measure of |
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In the figure shown above, if AB = DE, AC = DF and BC = EF, then which two triangles are congruent?
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| Right Option : C | |||
| View Explanation | |||
In the figure below,
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| Right Option : B | |||
| View Explanation | |||
In the given figure, AD is the median then ∠BAD is | |||
| Right Option : D | |||
| View Explanation | |||
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