
In quadrilateral $ ABCD $ , diagonals $ AC $ and $ BD $ intersect at point $ E $ such that $ AE:EC = BE:ED $ .
Show that: $ ABCD $ is a trapezium.
Answer
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Hint: In this question, we need to prove $ ABCD $ is a trapezium. Here, we will construct a quadrilateral $ ABCD $ with diagonals $ AC $ and $ BD $ intersect at point $ E $ . Then, using the properties of basic proportionality theorem, vertically opposite angles, SAS similarity theorem, corresponding angles congruence we will determine $ ABCD $ is a trapezium.
Complete step-by-step answer:
Now, we have constructed a quadrilateral $ ABCD $ with diagonals $ AC $ and $ BD $ intersect at point $ E $ .
Then, $ \dfrac{{AE}}{{EC}} = \dfrac{{BE}}{{ED}} $
$ \dfrac{{AE}}{{BE}} = \dfrac{{EC}}{{ED}} $
Let this be equation (1),
In $ \Delta ABE $ and $ \Delta CDE $ ,
Therefore, from equation (1),
$ \dfrac{{AE}}{{BE}} = \dfrac{{EC}}{{ED}} $
We know that vertically opposite angles are equal. Therefore, we have,
$ \angle ABE = \angle DEC $
Then, we also know that, the SAS similarity theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.
Therefore, by the theorem, we have,
$ \Delta ABE \sim \Delta CDE $
Here, the corresponding angles are equal,
$ \angle EDC = \angle EBA $
Therefore,
$ \angle BDC = \angle ABD $
From the figure we can say this is a pair of alternate angles.
Thus, $ AB\parallel DC $
Hence, the quadrilateral $ ABCD $ is trapezium.
So, the correct answer is “ $ ABCD $ is trapezium”.
Note: It is important to note here that vertically opposite angles are the angles opposite to each other when two lines cross. These angles are equal. Also, the angle rule of corresponding angles states that the corresponding angles are equal if a transversal cuts two parallel lines. However, we can also prove this by constructing a line $ EF $ where $ F $ in the side $ AD $ . Then, we can use basic proportionality theorem and converse of basic proportionality theorem to prove $ ABCD $ is a trapezium.
Complete step-by-step answer:
Now, we have constructed a quadrilateral $ ABCD $ with diagonals $ AC $ and $ BD $ intersect at point $ E $ .
Then, $ \dfrac{{AE}}{{EC}} = \dfrac{{BE}}{{ED}} $
$ \dfrac{{AE}}{{BE}} = \dfrac{{EC}}{{ED}} $
Let this be equation (1),
In $ \Delta ABE $ and $ \Delta CDE $ ,
Therefore, from equation (1),
$ \dfrac{{AE}}{{BE}} = \dfrac{{EC}}{{ED}} $
We know that vertically opposite angles are equal. Therefore, we have,
$ \angle ABE = \angle DEC $
Then, we also know that, the SAS similarity theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.
Therefore, by the theorem, we have,
$ \Delta ABE \sim \Delta CDE $
Here, the corresponding angles are equal,
$ \angle EDC = \angle EBA $
Therefore,
$ \angle BDC = \angle ABD $
From the figure we can say this is a pair of alternate angles.
Thus, $ AB\parallel DC $
Hence, the quadrilateral $ ABCD $ is trapezium.
So, the correct answer is “ $ ABCD $ is trapezium”.
Note: It is important to note here that vertically opposite angles are the angles opposite to each other when two lines cross. These angles are equal. Also, the angle rule of corresponding angles states that the corresponding angles are equal if a transversal cuts two parallel lines. However, we can also prove this by constructing a line $ EF $ where $ F $ in the side $ AD $ . Then, we can use basic proportionality theorem and converse of basic proportionality theorem to prove $ ABCD $ is a trapezium.
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