question_answer

\[\Delta ABC\] is a right triangle, right-angled at A and \[AD\bot BC\]. Then, \[\frac{BD}{DC}\] is equal to:

A) \[{{\left( \frac{AB}{AC} \right)}^{2}}\]

B) \[\frac{AB}{AC}\]

C) \[{{\left( \frac{AB}{AD} \right)}^{2}}\]

D) \[\frac{AB}{AD}\]

Correct Answer: A

Solution :

[a] In a right triangle ABC, right-angled at A and \[AD\bot BC\].

In \[\Delta CAB\] and \[\Delta CDA,\]

\[\angle CAB=\angle CDA\]            (Each\[90{}^\circ \])

\[\angle ACB=\angle DCA\]            (Common)

\[\therefore \,\,\Delta CAB\tilde{\ }\Delta CDA\](By AA similarity criterion)

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{BC}{AC}=\frac{AC}{DC}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,A{{C}^{2}}=BC\cdot DC\]                        ..(1)

Similarly,  \[\Delta CAB\tilde{\ }\Delta ADB\]

(By AA similarity criterion)

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{AB}{BD}=\frac{BC}{AB}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A{{B}^{2}}=BC\cdot BD\]  .(2)

Dividing (2) by (1), we set

\[\frac{A{{B}^{2}}}{A{{C}^{2}}}=\frac{BC\cdot BD}{BC\cdot DC}=\frac{BD}{DC}\]