Question

In ∆ABC, D and E are points on AB and AC respectively, DE is parallel

to BC. Prove that :
AD/AE
=
AB/AC​

Answer 1

Answer:

In order to prove that the points B,C,E and D are concyclic, it is sufficient to show that ∠ABC+∠CED=180$0 and ∠ACB+∠BDE=180

0

.

In △ABC, we have

AB=AC and AD=AE

⇒ AB−AD=AC−AE

⇒ DB=EC

Thus, we have

AD=AE and DB=EC

⇒

DB

AD

=

EC

AE

⇒ DE∣∣BC [By the converse of Thale's Theorem]

⇒ ∠ABC=∠ADE [Corresponding angles]

⇒ ∠ABC+∠BED=∠ADE+∠BDE [Adding ∠BDE both sides]

⇒ ∠ABC+∠BDE=180

0

⇒ ∠ACB+∠BDE=180

0

[∵AB=AC∴∠ABC=∠ACB]

Again, DE∣∣BC

⇒ ∠ACB=∠AED

⇒ ∠ACB+∠CED=∠AED+∠CED [Adding ∠CE on both sides]

⇒ ∠ACB+∠CED=180

0

⇒ ∠ABC+∠CED=180

0

[∵∠ABC=∠ACB]

Thus, BDEC is a cyclic quadrilateral. Hence, B,C,E and D are concyclic points.