A) \[A{{B}^{2}}+C{{D}^{2}}=B{{C}^{2}}+A{{D}^{2}}\]
B) \[C{{D}^{2}}+B{{D}^{2}}=2\,A{{D}^{2}}\]
C) \[A{{B}^{2}}+A{{C}^{2}}=2\,A{{D}^{2}}\]
D) \[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\]
Correct Answer: A
Solution :
[a] In \[\Delta ABC,\] \[A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}\] (i) In \[\Delta ACD,\] \[A{{D}^{2}}=A{{C}^{2}}+C{{D}^{2}}\] \[\Rightarrow \] \[A{{C}^{2}}=A{{D}^{2}}-C{{D}^{2}}\] (ii) From Eqs. (i) and (ii), we get \[A{{B}^{2}}=A{{D}^{2}}-C{{D}^{2}}+B{{C}^{2}}\] \[\Rightarrow \] \[A{{B}^{2}}+C{{D}^{2}}=A{{D}^{2}}=A{{D}^{2}}+B{{C}^{2}}\] |
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