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