In ΔABC, AB > AC; the bisector of ∠BAC meets BC at D. Let’s cut off a line segment AE equal in length of AC from the side AB. Let’s join D and E. Let’s prove that

(i) ΔACD≅ΔAED

(ii) ∠ACB > ∠ABC.

Given: ΔABC, AB > AC, the bisector of ∠BAC meets BC at D, AE = AC

To prove: (i) ΔACD≅ΔAED

(ii) ∠ACB > ∠ABC.

The diagram for the given criteria is as shown below,

(i) Consider ΔAED and ΔACD

AD = AD (common)

∠EAD = ∠CAD (as AD is the bisector of ∠BAC)

AE = AC (given)

Hence by SAS congruency criteria,

ΔAED≅ΔACD

Hence proved

(ii) In the given ΔABC,

∠ACB is opposite to side AB

And ∠ABC is opposite to the side AC

And given AB > AC

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠ACB > ∠ABC

Hence Proved