Question

In figure, $BA\perp AC,DE\perp EF$, such that $BA=DE$ and $BF=CD$, prove that $AC=EF$.

Answer 1

To prove: $AC=EF$, we have to prove
$\Delta ABC=\Delta DEF$
Here $BA=DE$ (given) …(i)
$BA\perp AC$ and $DE\perp FE$
$\angle BAC=\angle DEF={90}^0$ (each) …(ii)
Also $BF=CD$ (given) …(iii)
Adding $FC$ in equation (iii), we get
$BF+FC=CD+FC$
$BC=FD$ …(iv)
From (i), (ii) and (iv), we get
$\Delta ABC=\Delta DEF$ (by RHS congruency property)
$AC=EF$ (by c.p.c.t)
Hence proved.