Theorem: Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.
Given: , in which
To Prove:
Proof: On comparing the sides AC and DE of , there are three possibilities:
(i) AC = DE (II) AC < DE (iii) AC > DE
Case (i): when AC = DE, then in AC = DE [ Assumed] [ Given ] CB = EF [Given] Hence, [ By SAS Criterion ] |
Case (ii) : When AC < DE, then we take point G on DE such that AC = GE , Join GF Now in AC = GE [ Assumed] CB = EF [ Given ] [Given ] Hence, [ By SAS Criterion ] Hence [CPCT] .............(1) But [ Given ] .............(2) Therefore From (1) and (2) we get But this is possible only if G and D coincide Therefore AC = DE , Hence [ By SAS Criterion] |
Case (iii): When AC > DE, then we take point G on AC such that GC = DE
Then as in Case (ii) we can prove that A coincide with G i.e. AC = DE, Hence [ By SAS Criterion]
Hence in all the three cases
Illustration: AB is a line segment, AX and BY are two equal line segments drawn on opposite sides of line AB such that . If AB and XY intersect each other at P , prove that
(i) (ii) AB and XY bisect each other.
Solution: Since and transversal AB intersects them at A and B respectively. Therefore, [Alternate angles] Similarly, we have [ Transversal XY intersects parallel lines AX and BY at X and Y respectively] Thus, in triangles PAX and PBY, we have
[Given] and, AX = BY So, by ASA congruence criterion, we have
AP = BP and PX = PY Hence, and AB and XY bisect each other. |
Illustration: AB is line segment and P is its mid - point. D and E are points on the same side of AB such that
BAD = ABE and EPA = DPB.Show that
(i) DAP EBP
(ii) AD = BE
Solution:
Given: AB is a line segment and P is its mid - points. D and E are points on the same side of AB such that BAD = EPA = DPB.
Prove that : (i) DAP EBP
(ii) AD = BE
Proof: (i) Since, P is the mid - point of the line segment AB
In DAP and EBP,
AP = BP
Also, given DAP = EBP
and EPA = DPB
Adding EPD to both sides
EPA + EPD = EPD + DPB
APD = BPE
Thus, by ASA rule
DAPEBP
(ii) Since, DAP EBP (From above)
AD = BE (CPCT)
In two triangles, ABC and PQR, ∠A = 30°, ∠B = 70°, ∠P = 70°, ∠Q = 80° and AB = RP, then | |||
Right Option : C | |||
View Explanation |
In the below figure, if EF = QR then the congruence rule used for the congruency of the given triangles is ________________ | |||
Right Option : B | |||
View Explanation |
â–³ABC and â–³PQR are congruent under the correspondence ABC ↔ RQP. Write the parts of ΔABC that correspond to RP. | |||
Right Option : A | |||
View Explanation |
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