Theorem: Two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.
Given: Two triangles ABC and DEF such that To Prove: Proof: Place such that vertex A falls on vertex D and the side AB falls on side DE, As AB = DE [Given] So vertex B falls on vertex E. Since . Therefore, AC will fall on DF. But AC = DF [ Given] Therefore, C will fall on F. Thus, AC coincides with DF. Now, B falls on E and C falls on F. Therefore, BC coincides with EF. Thus, will coincide with . Hence, by definition of congruence, |
Illustration: In and , AB = PQ, BC = QR and CB and RQ are extended to X and Y respectively and Prove that |
Solution: We have,
[ Similarly, ]
.....(i)
In and we have
AB = PQ [Given]
[From (i)]
and, BC = QR [Given]
So, by SAS criterion of congruence, we have
Illustration: In the given figure, AC = AE, AB = AD and .Show that BC = DE
Solution: We have,
[Given]
[Adding to both sides]
Now, in triangles ABC and ADE,
[Given]
[From (i)]
and, [Given]
So, by SAS congruence criterion, we have
[CPCT]
In the given figure, and .If then which congruency rule is used to prove this? | |||
Right Option : C | |||
View Explanation |
In the figure given below, X and Y are two points on equal sides AB and AC of a such that AX = AY. Then by which criteria? | |||
Right Option : A | |||
View Explanation |
In the given figure, AB = AC. If , then measure of is | |||
Right Option : B | |||
View Explanation |
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