Congruent triangle

[darkyadoo]

Goog evening

Suppose two triangles [imath]\triangle ABC[/imath]and [imath]\triangle A'B'C'[/imath]

The Side-Angle-Side theorem says

if [imath]AB=A'B', \;AC=A'C',\;\angle A\cong\angle A'[/imath] then [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

My first question is the following one : Is that a theorem or an axiom?

Now let's go on the Angle-Side-Angle theorem ( ASA) which says :

if [imath]\angle A\cong \angle A', \, \angle B\cong B', AB=A'B'[/imath] then [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

Is my proof correct?

Proof:
First step we show that [imath]\overline {BC}\cong \overline {B'C'}[/imath] and then we apply the S-A-S theorem

We suppose that [imath]BC\ne B'C'[/imath], and let [imath]D[/imath] a point on the straightline [imath](B'C')[/imath] such that [imath]BC=B'D[/imath], regarding the S-A-S theorem [imath]\triangle ABC\cong\triangle A'B'C'[/imath], therefore [imath]\angle A\cong \angle DA'B'[/imath] so [imath]\angle DA'B'\cong \angle A'[/imath] since [imath]\angle A\cong \angle A'[/imath], so we have a contradiction therefore [imath]BC= B'C'[/imath].
As [imath]BC= B'C'[/imath],[imath]AB= A'B'[/imath] and [imath]\angle B\cong B'[/imath] from the S-A-S theorem we conclude that [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

 

[darkyadoo]

a mistake in my proof : [...]such that [imath]BC=B'D[/imath], regarding the S-A-S theorem [imath]\textcolor{red}{\triangle ABC\cong \triangle A'B'D}[/imath], therefore [imath]\angle A\cong \angle DA'B'[/imath] [...]

 

[Dr.Peterson]

Goog evening

Suppose two triangles [imath]\triangle ABC[/imath]and [imath]\triangle A'B'C'[/imath]

The Side-Angle-Side theorem says

if [imath]AB=A'B', \;AC=A'C',\;\angle A\cong\angle A'[/imath] then [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

My first question is the following one : Is that a theorem or an axiom?

That depends on your textbook. Some may take this as an axiom, while others may prove it from a different axiom. Ultimately, it doesn't matter.

Now let's go on the Angle-Side-Angle theorem ( ASA) which says :

if [imath]\angle A\cong \angle A', \, \angle B\cong B', AB=A'B'[/imath] then [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

Is my proof correct?

Proof:
First step we show that [imath]\overline {BC}\cong \overline {B'C'}[/imath] and then we apply the S-A-S theorem

We suppose that [imath]BC\ne B'C'[/imath], and let [imath]D[/imath] a point on the straightline [imath](B'C')[/imath] such that [imath]BC=B'D[/imath], regarding the S-A-S theorem [imath]\triangle ABC\cong\triangle A'B'C'[/imath], therefore [imath]\angle A\cong \angle DA'B'[/imath] so [imath]\angle DA'B'\cong \angle A'[/imath] since [imath]\angle A\cong \angle A'[/imath], so we have a contradiction therefore [imath]BC= B'C'[/imath].
As [imath]BC= B'C'[/imath],[imath]AB= A'B'[/imath] and [imath]\angle B\cong B'[/imath] from the S-A-S theorem we conclude that [imath]\triangle ABC\cong\triangle A'B'C'[/imath]

After "regarding the S-A-S theorem", did you mean to say [imath]\triangle ABC\cong\triangle A'B'D[/imath], rather than [imath]\triangle ABC\cong\triangle A'B'C'[/imath]?

 

[darkyadoo]

yes, exactly