Question

in the given figure BA is perpendicular to AC, DE is perpendicular to DF such that BA =DE and BF=EC. Show that triangle ABC is congruent to triangle DEF​

Answer 1

Given :

BA ⊥ AC, DE ⊥ DF and BA = DE, BF= EC

To Find :

We have to prove that ΔABC ≅ ΔDEF.

Solution :

As shown in the figure, BF = EC

On adding CF on both sides, we get

                                BF+CF = EC+CF

⇒                                    BC = EF      

RHS Congruence Theorem: In two right-angled triangles, if the length of one side of the triangle and the hypotenuse, is equal to the length of the corresponding side of the other triangle and its hypotenuse, then the two triangles are congruent.

In ΔABC and ΔDEF,

                                 ∠A = ∠D = 90°                               ∵BA ⊥ AC, DE ⊥ DF

                                       BC = EF

                                       BA = DE                                    ∵ given

∴ By RHS congruence rule, ΔABC ≅ ΔDEF.

  Hence proved.