A) \[2DC\cdot AD\]
B) \[2AD\cdot BC\]
C) \[3DC\cdot AD\]
D) \[\frac{1}{2}AD\cdot DC\]
Correct Answer: A
Solution :
As ADB is a right angled triangle. So, \[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\] \[\Rightarrow \] \[A{{C}^{2}}=A{{D}^{2}}+B{{D}^{2}}\] \[(\because AB=AC)\] \[\Rightarrow \]\[{{(AD+DC)}^{2}}=A{{D}^{2}}+B{{D}^{2}}\] \[\Rightarrow \]\[A{{D}^{2}}+D{{C}^{2}}+2AD\cdot DC=A{{D}^{2}}+B{{D}^{2}}\] \[\therefore \] \[B{{D}^{2}}-C{{D}^{2}}=2CD\cdot AD\]
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