Question

Prove that two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of another triangle.

Answer 1

Hint: There are three cases possible in such condition when in two triangles ABC and DEF given, \[\angle {\text{B = }}\angle {\text{E}}\] and $\angle {\text{C = }}\angle {\text{F}}$ and BC=EF - (i) If we assume side AB=DE then we can prove the triangles are congruent by SAS (Side-Angle-Side) rule (ii) If we assume that side AB>DE then we can take a point X on side AB such that XB=DE. Now we can prove that the two triangles are congruent if point X coincides with A. (iii) if we assume Side AB
Complete step-by-step answer:
Let us consider, ABC and DEF are such two triangles in which given,
\[\angle {\text{B = }}\angle {\text{E}}\] And $\angle {\text{C = }}\angle {\text{F}}$ and BC=EF. Now we have to prove that the triangles are congruent, $\Delta {\text{ABC}} \cong \Delta {\text{DEF}}$.Now there are three cases possible,
(i)Let us assume that side AB=DE,

Now, we are given that BC=EF and \[\angle {\text{B = }}\angle {\text{E}}\]and AB=DE (assumed) then by SAS axiom we can prove that $\Delta {\text{ABC}} \cong \Delta {\text{DEF}}$
(ii) Let us assume that AB>DE then
Construction- Locate point X inside AB of triangle ABC such that XB=DE.

Now, in $\Delta {\text{XBC}}$ and $\Delta {\text{DEF}}$,
XB=DE (from construction)
\[\angle {\text{B = }}\angle {\text{E}}\] (Given)
BC=EF (given)
So, by SAS axiom $\Delta {\text{XBC}} \cong \Delta {\text{DEF}}$
Since the triangles are congruent then their corresponding parts are also equal. So, $\angle {\text{PCB = }}\angle {\text{DFE}}$ and we are given that $\angle {\text{XCB = }}\angle {\text{DFE}}$ so if the point P coincides with A then we can also say that $\angle {\text{XCB = }}\angle {\text{ACB}}$.Then by SAS axiom we can prove that , $\Delta {\text{ABC}} \cong \Delta {\text{DEF}}$
(iii) Let us assume that side ABConstruction-Construct a point Y in side DE such that AB=YE

Now, in $\Delta {\text{YEF}}$ and $\Delta {\text{ABC}}$ ,
YE=AB (from construction)
\[\angle {\text{B = }}\angle {\text{E}}\] (Given)
BC=EF (given)
So, by SAS axiom $\Delta {\text{YEF}} \cong \Delta {\text{ABC}}$
Since the triangles are congruent then their corresponding parts are also equal. So, $\angle {\text{ACB = }}\angle {\text{YFE}}$ and we are given that $\angle {\text{ACB = }}\angle {\text{DFE}}$ so if the point Y coincides with D then we can also say that$\angle {\text{DFE = }}\angle {\text{YFE}}$ .Then by SAS axiom we can prove that , $\Delta {\text{ABC}} \cong \Delta {\text{DEF}}$
Thus Angle-Side-Angle Congruence is proved.

Note: We have used SAS congruence rule to prove ASA congruence rule in the given question.SAS axiom or congruence rule states that two triangles are congruent if two sides and included angle of one triangle are equal to the two sides and included angle of other triangle.