# In the given figure, ABCD is a trapezium with ${\text{AB}}\ parallel {\text{DC}}$. If $\Delta AED$ is similar to $\Delta BEC$. Prove that AD = BC

Last updated date: 25th Mar 2023

•

Total views: 310.5k

•

Views today: 7.87k

Answer

Verified

310.5k+ views

Hint- Check for the relation between corresponding angles and sides of both triangles.

Given: \[ABCD\] is a trapezium with ${\text{AB}}\parallel {\text{DC}}$.

Diagonals $AC$ and $BD$ intersect at point $E$.

To prove: ${\text{AD = BC}}$

Now, $\Delta AED \sim \Delta BEC{\text{ }}\left( {\because {\text{Given}}} \right)$

$\therefore \dfrac{{{\text{AE}}}}{{{\text{BE}}}}{\text{ = }}\dfrac{{{\text{ED}}}}{{{\text{EC}}}}{\text{ = }}\dfrac{{{\text{AD}}}}{{{\text{BC}}}}{\text{ }} \ldots \ldots \left( 1 \right)\left( {{\text{corresponding sides are proportional}}} \right)$

In \[\Delta {\text{ABE}}\]and \[\Delta {\text{CDE}}\]:

\[

\angle {\text{AEB = }}\angle {\text{CED }}\left( {{\text{vertically opposite angles}}} \right) \\

\angle {\text{EAB}} = {\text{ }}\angle {\text{ECD }}\left( {{\text{alternate angles}}} \right) \\

\]

Hence, by AA similarity, we can say

\[\Delta {\text{ABE}} \sim \Delta {\text{CDE }}\]

Now, \[\dfrac{{{\text{AB}}}}{{{\text{CD}}}} = \dfrac{{{\text{EB}}}}{{{\text{ED}}}} = \dfrac{{{\text{AE}}}}{{{\text{EC}}}}{\text{ }}\left( {\because {\text{corresponding sides are proportional}}} \right)\]

Therefore, \[\dfrac{{{\text{EC}}}}{{{\text{ED}}}} = \dfrac{{{\text{AE}}}}{{{\text{EB}}}}{\text{ }} \ldots \ldots \left( 2 \right)\]

From \[\left( 1 \right)\]and \[\left( 2 \right)\], we get

\[\dfrac{{{\text{AD}}}}{{{\text{BC}}}} = \dfrac{{{\text{ED}}}}{{{\text{EC}}}}\]

As we know, \[{\text{ED = EC}}\]

\[

\Rightarrow \dfrac{{{\text{AD}}}}{{{\text{BC}}}} = 1{\text{ }} \\

\therefore {\text{ AD = BC}} \\

\]

Hence Proved.

Note- Two triangles are similar if their corresponding angles are congruent and corresponding sides are proportional. There are many criterions of similarity and many properties of similar triangles which come handy while solving problems like these.

Given: \[ABCD\] is a trapezium with ${\text{AB}}\parallel {\text{DC}}$.

Diagonals $AC$ and $BD$ intersect at point $E$.

To prove: ${\text{AD = BC}}$

Now, $\Delta AED \sim \Delta BEC{\text{ }}\left( {\because {\text{Given}}} \right)$

$\therefore \dfrac{{{\text{AE}}}}{{{\text{BE}}}}{\text{ = }}\dfrac{{{\text{ED}}}}{{{\text{EC}}}}{\text{ = }}\dfrac{{{\text{AD}}}}{{{\text{BC}}}}{\text{ }} \ldots \ldots \left( 1 \right)\left( {{\text{corresponding sides are proportional}}} \right)$

In \[\Delta {\text{ABE}}\]and \[\Delta {\text{CDE}}\]:

\[

\angle {\text{AEB = }}\angle {\text{CED }}\left( {{\text{vertically opposite angles}}} \right) \\

\angle {\text{EAB}} = {\text{ }}\angle {\text{ECD }}\left( {{\text{alternate angles}}} \right) \\

\]

Hence, by AA similarity, we can say

\[\Delta {\text{ABE}} \sim \Delta {\text{CDE }}\]

Now, \[\dfrac{{{\text{AB}}}}{{{\text{CD}}}} = \dfrac{{{\text{EB}}}}{{{\text{ED}}}} = \dfrac{{{\text{AE}}}}{{{\text{EC}}}}{\text{ }}\left( {\because {\text{corresponding sides are proportional}}} \right)\]

Therefore, \[\dfrac{{{\text{EC}}}}{{{\text{ED}}}} = \dfrac{{{\text{AE}}}}{{{\text{EB}}}}{\text{ }} \ldots \ldots \left( 2 \right)\]

From \[\left( 1 \right)\]and \[\left( 2 \right)\], we get

\[\dfrac{{{\text{AD}}}}{{{\text{BC}}}} = \dfrac{{{\text{ED}}}}{{{\text{EC}}}}\]

As we know, \[{\text{ED = EC}}\]

\[

\Rightarrow \dfrac{{{\text{AD}}}}{{{\text{BC}}}} = 1{\text{ }} \\

\therefore {\text{ AD = BC}} \\

\]

Hence Proved.

Note- Two triangles are similar if their corresponding angles are congruent and corresponding sides are proportional. There are many criterions of similarity and many properties of similar triangles which come handy while solving problems like these.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE