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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