Question

In figure $3.51,$ in $\Delta ABC,$ seg $AD$ and seg $BE$ are altitudes and $AE\cong BD.$
Prove that seg $AD\cong$ seg $BE$

Answer 1

Let the point of inter-section of altitudes be $x,$

in $△DXB,$

$\angle EXA=\angle BXD$ (vertically opposite angles)

$\angle XEA=\angle XDB$ (Right angles)

and $AE=BD$

So, $△DXB\cong △EXA$

and thus, $EX=XD$ (CPCT) $⟶i$

and $XB=XA$ (CPCT) $⟶ii$

Adding $i$ and $ii$ , we get...

$BE=AD.$

Hence proved.