| In the figure, if \[\angle BAC=90{}^\circ \] and \[AD\bot BC,\] then: (NCERT EXEMPLAR) |
 |
A) \[BD.CD=B{{C}^{2}}\]
B) \[AB.AC=B{{C}^{2}}\]
C) \[BD.CD=A{{D}^{2}}\]
D) \[AB.AC=A{{D}^{2}}\]
Correct Answer: C
Solution :
| [c] In \[\Delta ABC,\] |
| \[\angle B+\angle BAC+\angle C=180{}^\circ \] |
| (By angle sum property) |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle B+90{}^\circ +\angle C=180{}^\circ \] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle B=90{}^\circ -\angle C\] .(1) |
| Similarly, In \[\Delta ADC,\] |
| \[\angle DAC=90{}^\circ -\angle C\] ..(2) |
| In \[\Delta ADB\] and \[\Delta CDA,\] |
| \[\angle ADB=\angle CDA\] (Each \[90{}^\circ \]) |
| \[\angle DBA=\angle DAC\] (From (1) and (2)] |
| \[\therefore \,\,\,\,\,\,\,\,\Delta ADB\tilde{\ }\Delta CDA\] |
| (By AA similarity criterion) |
| \[\therefore \,\,\,\,\,\,\,\frac{BD}{AC}=\frac{AD}{CD}\,\,\,\Rightarrow \,\,\,BD\cdot CD=A{{D}^{2}}\] |
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