Question
In the adjoining figure, ABC is a triangle in which AB = AC. If D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.
✓
Answer
Given:
AD = AE ...(i)
AB = AC ...(ii)
Subtracting AD from both sides, we get:
⇒ AB - AD = AC - AD
⇒ AB - AD = AC - AE (Since, AD = AE)
⇒ BD = EC ...(iii)
Dividing equation (i) by equation (iii), we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Dividing the converse pf Thales' theorem, DE || BC
$\Rightarrow\angle\text{DEC}+\angle\text{ECB}=180^\circ$ (Sum of interior angles on the same side of a transversal line is 180º)
$\Rightarrow\angle\text{DEC}+\angle\text{CBD}=180^\circ$ $($Since, AB = AC ⇒ $\angle\text{B}=\angle\text{C})$
Hence, quadrilateral BDED is cyclic.
Therefore, B, C, E and D are concyclic points.
AD = AE ...(i)
AB = AC ...(ii)
Subtracting AD from both sides, we get:
⇒ AB - AD = AC - AD
⇒ AB - AD = AC - AE (Since, AD = AE)
⇒ BD = EC ...(iii)
Dividing equation (i) by equation (iii), we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Dividing the converse pf Thales' theorem, DE || BC
$\Rightarrow\angle\text{DEC}+\angle\text{ECB}=180^\circ$ (Sum of interior angles on the same side of a transversal line is 180º)
$\Rightarrow\angle\text{DEC}+\angle\text{CBD}=180^\circ$ $($Since, AB = AC ⇒ $\angle\text{B}=\angle\text{C})$
Hence, quadrilateral BDED is cyclic.
Therefore, B, C, E and D are concyclic points.
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