AB is a straight line, C and D are points the same side of AB such that CA perpendicular to AB and DB is perpendicular to AB. Let AD and BC meet at E. What is \[\frac{AE}{AD}+\frac{BE}{BC}\] equal to? |
A) 2
B) 1.5
C) 1
D) None of these
Correct Answer: D
Solution :
Since, AB is a straight line and C and D are points such that \[AC\bot AB\] and \[BD\bot AB.\] |
\[\therefore \] \[AC\parallel BD\] |
So, ABCD forms trapezium. |
Now, by property of trapezium diagonals intersect each other in the ratio of lengths of parallel sides. |
\[\therefore \] \[\frac{AE}{ED}=\frac{BE}{CE}\] |
\[\frac{AE}{AD-AE}=\frac{BE}{BC-BE}\] |
\[\frac{BC-BE}{BE}=\frac{AD-AE}{AE}\] |
\[\frac{BC}{BE}-1=\frac{AD}{AE}-1\]\[\Rightarrow \]\[\frac{BC}{BE}=\frac{AD}{AE}\] |
\[\frac{AE}{AD}=\frac{BE}{BC}\] |
But the value of \[\frac{AE}{AD}\] or \[\frac{BE}{BC}\]cannot be determined. |
So, we cannot find the value of \[\frac{AE}{AD}+\frac{BE}{BC}.\] |
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