question_answer

In given figure, \[EB\bot AC,\text{ }BG\bot AE\] and \[CF\bot AE\] Prove that:

(i)\[\Delta \,ABG\sim \Delta \,DCB\]

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

Answer:

Given: \[EB\bot AC,\text{ }BG\bot AE\] and \[CF\bot AE\]

To prove: (i) \[\Delta \,ABG\sim \Delta \,DCB\]

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

Proof: (i) In \[\Delta \text{ }ABG\] and \[\Delta \text{ }DCB,\text{ }BG\parallel CF\] as corresponding angles are equal.

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

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

\[\therefore \Delta \text{ }ABG\sim \Delta \text{ }DCB\]                              Hence Proved.

[By AA similarity]

\[\angle 1=\angle 3\]                          [CPCT]

(ii) In \[\Delta \text{ }ABE\]and \[\Delta \text{ }DBC\]

\[\angle 1=\angle 3\]               [Proved above]

\[\angle ABE=\angle 5\]

[Each is \[90{}^\circ ,\text{ }EB\bot AC\](Given)]

\[\Delta \text{ }ABE\sim \Delta \text{ }DBC\]                      [By AA similarity]

In similar triangles, corresponding sides are proportional

\[\therefore \frac{BC}{BD}=\frac{BE}{BA}\]                   Hence Proved.