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D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
Circles | Class 10 | Exercise 17C | ICSE | Maths | Selina
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.

Given,

\[\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC\text{ }and\text{ }D\text{ }and\text{ }E\]are points on \[AB\text{ }and\text{ }AC\]such that \[AD\text{ }=\text{ }AE\]

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 8

And, \[DE\] is joined.

Required to prove: Points \[B,\text{ }C,\text{ }E\text{ }and\text{ }D\] are concyclic

Proof:

In \[\vartriangle ABC\]

\[AB\text{ }=\text{ }AC\] [Given]

So, \[\angle B\text{ }=\angle C\] [Angles opposite to equal sides]

Similarly,

In \[\vartriangle ADE\]

\[AD\text{ }=\text{ }AE\][Given]

So, \[\angle ADE\text{ }=\angle AED\][Angles opposite to equal sides]

Now, in \[\vartriangle ABC\]we have

\[AD/AB\text{ }=\text{ }AE/AC\]

Hence, \[DE\text{ }||\text{ }BC\][Converse of BPT]

So,

\[\angle ADE\text{ }=\angle B\][Corresponding angles]

\[({{180}^{o}}-\angle EDB)\text{ }=\angle B\]

\[\angle B\text{ }+\angle EDB\text{ }=\text{ }{{180}^{o}}\]

But, it’s proved above that

\[\angle B\text{ }=\angle C\]

So,

\[\angle C\text{ }+\angle EDB\text{ }=\text{ }{{180}^{o}}\]

Thus, opposite angles are supplementary.

Similarly,

\[\angle B\text{ }+\angle CED\text{ }=\text{ }{{180}^{o}}\]

Hence, \[B,\text{ }C,\text{ }E\text{ }and\text{ }D\]are concyclic.

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