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.
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