Question

In given figure, $EB\perp AC,BG\perp AE$ and $CF\perp AE$ Prove that:

(i) $\Delta ABG\sim \Delta DCB$
(ii) $\frac{BC}{BD}=\frac{BE}{BA}$

Answer 1

Given: $EB\perp AC,BG\perp AE\text{and}CF\perp AE$

To prove: (i) $\Delta ABG\sim \Delta DCB$

(ii) $\frac{BC}{BD}=\frac{BE}{BA}$

Proof: (i) In $\Delta ABG$ and $\Delta DCB$, $BG||CF$ as corresponding angles are equal.

$\angle 2=\angle 7=\angle 5$ [Each ${90}^{\circ }$]

$\angle 6=\angle 4$ [Corresponding angles]

$\therefore \Delta ABG\sim \Delta DCB$

[By AA similarity]
$\angle 1=\angle 3$ [CPST]
(ii) In $\Delta ABE$ and $\Delta DBC$
$\angle 1=\angle 3$ [Proved above]
$\angle ABE=\angle 5$
[Each is ${90}^{\circ }$, $EB\perp AC$ (Given)]
$\Delta ABE\sim \Delta DBC$ [By AA similarity]
In similar triangles, corresponding sides are proportional
$\frac{AB}{BD}=\frac{BE}{BC}$
$\therefore \frac{BC}{BD}=\frac{BE}{BA}$