| In the given figure, \[\angle BAC=90{}^\circ \] and \[AD\bot BC,\] then: |
 |
A) \[BD\times CD=B{{C}^{2}}\]
B) \[AB\times AC=B{{C}^{2}}\]
C) \[BD\times CD=A{{D}^{2}}\]
D) \[AB\times AC=A{{D}^{2}}\]
Correct Answer: C
Solution :
| [c] In \[\Delta ABC\] and \[\Delta DBA\] |
| \[\angle BAC=\angle BDA\] (Each \[90{}^\circ \]) |
| \[\angle ABC=\angle DBA\] (Common) |
| \[\therefore \,\,\Delta ABC\tilde{\ }\Delta DBA\] |
| (By AA similarity criterion) ...(1) |
| Similarly, \[\Delta ABC\tilde{\ }\Delta DAC\] |
| (By AA similarity criterion) ..,(2) |
| From (1) and (2), we have \[\Delta DBA\tilde{\ }\Delta DAC\] |
| \[\therefore \,\,\,\,\,\,\,\,\frac{BD}{AD}=\frac{AD}{CD}\,\,\,\Rightarrow \,\,\,A{{D}^{2}}=BD\times CD\] |
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