Question

prove that two Triangles are congruent if two angles and the included side of the one Triangles are equal to the two angles and the included side of the other triangle

Answer 1

the theorem is Angle Angle side congruency rule

Answer 2

$\huge\underline\bold\red{Answer:}$

Given: Two ∆s ABC and DFC such that

$\angle$B = $\angle$E, $\angle$C = $\angle$F and BC = EF.

To prove: ∆ABC $\cong$ ∆DEF

Proof: There are three cases:

_________________

Case 1: When AB = DE

in this case, we have

• AB = DE
• $\angle$B = $\angle$E
• BC = EF

So, by SAS - criterion of congruence, ∆ABC $\cong$ ∆DEF.

___________________

Case 2: When AB < ED

In this case take a point P on ED such that PE = AB. Join PF.

Now, in ∆ABC and ∆PEF, we have

• AB = PE
• $\angle$B=$\angle$E
• BC = EF

So, by SAS - criterion of congruence, ∆ABC $\cong$ ∆PEF.

• $\angle$ACB=$\angle$PEF
• $\angle$ACB=$\angle$DFE
• $\angle$PFE=$\angle$DFE

This is possible only when ray FP or P coincides with D. Therefore, AB must be equal to DE.

Thus, in ∆ABC and ∆DEF, we have

• AB = DE
• $\angle$B=$\angle$E
• BC = EF

So, by SAS - criterion of congruence, ∆ABC $\cong$ ∆DEF.

____________________

Case 3: When AB > ED

In this case, take a point P on ED produced such that EP = AB. Join PF. Now, proceeding exactly on the same lines as in case 2, we can prove that

∆ABC $\cong$ ∆DEF.

Hence, ∆ABC $\cong$ ∆ DEF.