A) \[B{{C}^{2}}+A{{D}^{2}}+2AB\cdot CD\]
B) \[A{{B}^{2}}+C{{D}^{2}}2AD\cdot BC\]
C) \[A{{B}^{2}}+C{{D}^{2}}+2AB\cdot CD\]
D) \[B{{C}^{2}}+A{{D}^{2}}+2BC\cdot AD\]
Correct Answer: A
Solution :
In \[\Delta ABD,\]\[\angle A\]is acute |
So, \[B{{D}^{2}}=A{{D}^{2}}+A{{B}^{2}}-2AB\cdot AQ\] ...(i) |
In\[\Delta ABC,\]\[\angle B\]is cute |
So, \[A{{C}^{2}}=B{{C}^{2}}+A{{B}^{2}}-2AB\cdot AD\] ...(ii) |
Adding Eqs. (i) and (ii), we get |
\[\therefore \]\[A{{C}^{2}}+B{{D}^{2}}=(B{{C}^{2}}+A{{D}^{2}})+2AB(AB-BP-AQ)\] |
\[=(B{{C}^{2}}+A{{D}^{2}})+2AB\cdot PQ\] |
\[=B{{C}^{2}}+A{{D}^{2}}+2AB\cdot CD\] \[(\because PQ=DC)\] |
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