Hushar Mulga
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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.

Solution

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

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