Question

Q17. In AABC, D and E are points on AB and AC respectively such that DE is parallel to BC. If BD =ce
Prove that ABC is an isosceles triangle.​

Answer 1

Answer:

We have, DE || BC Therefore, by BPT, we have, Adding DB on both sides ⇒ AD + DB = AE + DB ⇒ AD + DB = AE + EC [∴ BD = CE] ⇒ AB = AC ⇒ Δ ABC is isoscelesRead more on Sarthaks.com - https://www.sarthaks.com/33493/if-d-and-e-are-points-on-sides-ab-and-ac-respectively-of-a-abc-such-that-de-bc