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.

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.