A) \[B{{C}^{2}}+A{{D}^{2}}+2AB\,\,.\,\,CD\]
B) \[A{{B}^{2}}+C{{D}^{2}}+2AD\,\,.\,\,BC\]
C) \[A{{B}^{2}}+C{{D}^{2}}+2AB\,\,.\,\,AD\]
D) \[B{{C}^{2}}+A{{D}^{2}}+2BC\,\,.\,\,AD\]
Correct Answer: A
Solution :
In \[\Delta \] ABD. \[\angle \]A is acute. So. \[B{{D}^{2}}=A{{D}^{2}}+A{{B}^{2}}-2AB\text{ }.\text{ }AQ\] ....(i) In \[\Delta \] ABC, \[\angle \]B is acute. So. \[A{{C}^{2}}=B{{C}^{2}}+A{{B}^{2}}-2AB.\text{ }AD\] ......(ii) Adding (i) and (ii) \[\therefore A{{C}^{2}}+B{{D}^{2}}=\left( B{{C}^{2}}+\text{ }A{{D}^{2}} \right)\] \[+\text{ }2AB\left( AB-BP-AQ \right)\] \[=\left( B{{C}^{2}}+A{{D}^{2}} \right)+2AB.\text{ }PQ\] \[=B{{C}^{2}}+A{{D}^{2}}+\text{ }2AB\text{ }.CD\] \[\text{ }\!\![\!\!\text{ }\because PQ=DC]\]You need to login to perform this action.
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