In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.
Practice Set 1.2 | Q 11 | Page 15
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.
In △ABC, ∠ABD = ∠DBC
\[\frac{AD}{DC} = \frac{AB}{BC} . . . \left( I \right) \left( \text{ By angle bisector theorem } \right)\]
In △ABC, ∠BCE = ∠ACE
\[\frac{AE}{EB} = \frac{AC}{BC} . . . \left( II \right) \left( \text{ By angle bisector theorem } \right)\]
From (1) and (2)
\[\frac{AD}{DC} = \frac{AE}{EB} \left( \because seg AB \cong seg AC \right)\] ED || BC (Converse of basic proportional theorem)
Answer:-
Given: In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB, and seg AB ≅ seg AC.
To prove: ED || BC.
Proof:
In △ABC, by the angle bisector theorem, we have:
∠ABD = ∠DBC and AD/DC = AB/BC …… (1)
∠BCE = ∠ACE and AE/EB = AC/BC …… (2)
Since seg AB ≅ seg AC, we have AB/BC = AC/BC, so AD/DC = AE/EB.
Now, let F be the intersection of ray BD and ray CE. By the angle bisector theorem, we have:
AF/FC = AB/BC = AC/BC = AE/EB.
Thus, △AFE is similar to △BCE by the side-angle-side (SAS) similarity criterion.
Therefore, ∠AED = ∠BCE = ∠ACE, and ED || BC by the converse of the basic proportionality theorem.
Hence, ED || BC.
Chapter 1. Similarity- Practice Set 1.2 – Page 15
Click Here for All Textbook Soutions of Chapter 1: Similarity