Question

AB and AC are two equal side of the triangle ABC on which two points D and E respectively. are such that AD = AE prove that B,C,E and D cyclic?

Answer 1

Given that AB = AC. Therefore ABC is an isosceles triangle.

Also given that AD = AE.

We need to prove that B,C,D and E are concyclic.

Since AB = AC and AD = AE, we have BD = DE.

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel tothe third side.

Thus, the line DE is parallel to the side BC.

In triangle ABC, since AB = AC, we have

In  triangle ADE, since AD = AE, we have

Thus in triangle ABC and ADE, we have,

and

Using equations (1) and (2), the above equations become

and

If the sum of any pair of opposite angles of a quadrilateral is 180 degrees, then the quadrilateral is cyclic.

Since the anglesare opposite angles of the quadrialteral BCED, then the quadrilateral is cyclic

.

Answer 2

Answer:

Step-by-step explanation:

Given :-

AB = AC

AD = AC

To Prove :-

Points B, C, E and D lie on a same circle.​

Solution :-

Since, AB = AC .....(i) and AD = AC...... (ii)

Subtracting AD from both sides, we get:  

⟹ AB - AD = AC - AD  

⟹ BD = EC …(iii)  (Since, AD = AE)  

Dividing equation (ii) by equation (iii), we get:

AD/DB = AE/EC

Applying the converse of Thales’ theorem, DE‖BC  

⟹ ∠DEC + ∠ECB = 180°

⟹ ∠DEC + ∠CBD = 180° (Since, AB = AC ⇒ ∠B = ∠C)  

Therefore, quadrilateral BCED is cyclic.

Hence, B,C,E and D are concylic points.