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In Figure, BA \perp AC, DE \perp DF such that BA = DE and BF = EC. Show that \triangleABC \cong \triangleDEF.

Answers (1)

Given, BA \perp AC

DE\perp DF

BA = DE and BF = EC

To show :- \triangle ABC \cong   \triangle DEF

Proof :-

BF = EC                                                          (given)

BF + FC = EC + FC                            (adding FC to LHS and RHS)

BC = FE

Also,

\angle CAB = \angle FDE = 90o                         (BA \perp AC, DE \perp DF given)

AB = DE                                                         (given)

By RHS criterion of congruence.

\triangle ABC \cong  \triangle DEF

Hence proved.

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