question_answer

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 \[\frac{AE}{AD}+\frac{BE}{BC}.\]