A) 8
B) 4
C) 3
D) 6
Correct Answer: D
Solution :
[d] AD extended meets BC at F. \[\angle \,ADB=\angle \,BDF=90{}^\circ \] \[\angle \,ADB=\angle \,FDB\](BD is the angle bisector) \[\therefore \] \[\angle \,BAD=\angle \,BFD\] \[\Rightarrow \] \[\Delta \,ABD\] and \[\Delta \,FBD\]are congruent. \[\Rightarrow \] AD = DF and \[\Delta \,ADE\]is similar to \[\Delta \,AFC\]\[(\therefore DE\parallel BC)\] \[\frac{AE}{AC}=\frac{AD}{AF}=\frac{1}{2}\]\[\Rightarrow \]\[AE=\frac{1}{2}\times 12=6\,cm\] |
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