Theorem: Two right triangles that have a congruent hypotenuse and a corresponding, congruent leg are congruent triangles.
Prove: Let ABC and DEF are the two right triangles. ( shown in below diagram)
Such that, AB =DE, AC = DF and ∠ABC = ∠ DEF = 90°
Prove: Δ ABC ≅ Δ DEF
Since, In Δ ABC,
[tex]AC^2 = AB^2 + BC^2[/tex] ( by Pythagoras theorem)
[tex]AC = \sqrt{AB^2 + BC^2}[/tex]
Similarly, In triangle DEF,
[tex]DF = \sqrt{DE^2 + EF^2}[/tex]
But, AC= DF ( given)
Therefore, [tex]\sqrt{AB^2 + BC^2}= \sqrt{DE^2 + EF^2}[/tex]
⇒ [tex]AB^2 + BC^2= DE^2 + EF^2[/tex]
⇒ [tex]DE^2 + BC^2=DE^2 + EF^2[/tex] ( AB= DE)
⇒ [tex]BC^2= EF^2[/tex]
⇒ BC= EF
Therefore, By SSS postulate of congruent,
Δ ABC ≅ Δ DEF
