
In the given figure ${\text{ABC}}$ is a triangle in which ${\text{AB = AC}}$. Points ${\text{D}}$ and ${\text{E}}$ are points on the sides ${\text{AB}}$and ${\text{AC}}$ respectively such that ${\text{AD = AE}}$. Show that the points ${\text{B,C,E}}$and ${\text{D}}$ are concyclic.
Answer
630.3k+ views
Hint: - Here, we go with the property of parallel lines and with the property of cyclic quadrilaterals. In a cyclic quadrilateral, the sum of either pair of opposite angles is supplementary.
In order to prove that the points ${\text{B,C,E}}$and ${\text{D}}$ are concyclic, it is sufficient to show that \[\angle ABC + \angle CED = {180^0}\] and \[\angle ACB + \angle BDE = {180^0}\].
In\[\;\vartriangle ABC\], we have
\[AB = AC\]And \[AD = AE\]
\[
\Rightarrow \;AB - AD = AC - AE \\
\Rightarrow \;DB = EC \\
\]
Thus, we have
\[AD = AE\;\]And \[DB = EC\;\]
\[ \Rightarrow \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\]
\[ \Rightarrow \;{\text{ }}\;DE\mid \mid BC\;{\text{ }}\] [By the converse of Thales Theorem]
\[ \Rightarrow \;{\text{ }}\;\angle ABC = \angle ADE\;{\text{ }}\;{\text{ }}\] [Corresponding angles]
\[ \Rightarrow \;{\text{ }}\;\angle ABC + \angle BDE = \angle ADE + \angle BDE\] [Adding \[\angle BDE\]both sides]
\[ \Rightarrow \;\;\angle ABC + \angle BDE = {180^0}{\text{ }}\] \[\because \angle BDE + \angle ADE = {180^0}\](Straight angle)
\[ \Rightarrow \;{\text{ }}\;\angle ACB + \angle BDE = {180^0}\] [\[\because AB = AC\]And\[\therefore \angle ABC = \angle ACB\]]
Again, \[DE\mid \mid BC\]
\[ \Rightarrow \;{\text{ }}\;\;\angle ACB = \angle AED\]
\[ \Rightarrow \;{\text{ }}\;\angle ACB + \angle CED = \angle AED + \angle CED\;\] [Adding \[\angle CED\]both sides]
\[ \Rightarrow \;\;\angle ACB + \angle CED = {180^0}\] \[\because \angle AED + \angle CED = {180^0}\](Straight angle)
\[ \Rightarrow \;\;\angle ABC + \angle CED = {180^0}\] [\[\because \angle ABC = \angle ACB\]]
Thus, \[BDEC\] is a cyclic quadrilateral because the sum of alternate angles are supplementary Hence ${\text{B,C,E}}$and ${\text{D}}$ are concyclic points.
Note:-This question is based on the property of parallel lines and their corresponding angles and also on the property of the isosceles triangle. By applying these properties we can easily solve such types of questions for showing cyclic quadrilaterals
In order to prove that the points ${\text{B,C,E}}$and ${\text{D}}$ are concyclic, it is sufficient to show that \[\angle ABC + \angle CED = {180^0}\] and \[\angle ACB + \angle BDE = {180^0}\].
In\[\;\vartriangle ABC\], we have
\[AB = AC\]And \[AD = AE\]
\[
\Rightarrow \;AB - AD = AC - AE \\
\Rightarrow \;DB = EC \\
\]
Thus, we have
\[AD = AE\;\]And \[DB = EC\;\]
\[ \Rightarrow \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\]
\[ \Rightarrow \;{\text{ }}\;DE\mid \mid BC\;{\text{ }}\] [By the converse of Thales Theorem]
\[ \Rightarrow \;{\text{ }}\;\angle ABC = \angle ADE\;{\text{ }}\;{\text{ }}\] [Corresponding angles]
\[ \Rightarrow \;{\text{ }}\;\angle ABC + \angle BDE = \angle ADE + \angle BDE\] [Adding \[\angle BDE\]both sides]
\[ \Rightarrow \;\;\angle ABC + \angle BDE = {180^0}{\text{ }}\] \[\because \angle BDE + \angle ADE = {180^0}\](Straight angle)
\[ \Rightarrow \;{\text{ }}\;\angle ACB + \angle BDE = {180^0}\] [\[\because AB = AC\]And\[\therefore \angle ABC = \angle ACB\]]
Again, \[DE\mid \mid BC\]
\[ \Rightarrow \;{\text{ }}\;\;\angle ACB = \angle AED\]
\[ \Rightarrow \;{\text{ }}\;\angle ACB + \angle CED = \angle AED + \angle CED\;\] [Adding \[\angle CED\]both sides]
\[ \Rightarrow \;\;\angle ACB + \angle CED = {180^0}\] \[\because \angle AED + \angle CED = {180^0}\](Straight angle)
\[ \Rightarrow \;\;\angle ABC + \angle CED = {180^0}\] [\[\because \angle ABC = \angle ACB\]]
Thus, \[BDEC\] is a cyclic quadrilateral because the sum of alternate angles are supplementary Hence ${\text{B,C,E}}$and ${\text{D}}$ are concyclic points.
Note:-This question is based on the property of parallel lines and their corresponding angles and also on the property of the isosceles triangle. By applying these properties we can easily solve such types of questions for showing cyclic quadrilaterals
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

What is the Full Form of ICSE / ISC ?

What is pollution? How many types of pollution? Define it

Explain the importance of pH in everyday life class 9 chemistry CBSE

