Question
In the we have $\angle\text{ABC}=\angle\text{ACB},\angle3=\angle4$ Show that $\angle1=\angle2$ 
✓
Answer
Given, $\angle\text{ABC}=\angle\text{ACB}\ ...(\text{i})$ And $\angle4=\angle3\ ...(\text{iii})$
According to Eulid’s axiom, if equals are subtracted from equals, then remainders are also equal. On subtracting Eq. $(ii)$ from Eq. $(i),$
we get $\angle\text{ABC}-\angle4=\angle\text{ACB}-\angle3$
$\Rightarrow\angle1=\angle2$
Now, in $ABCD$, $\angle1=\angle2$
$\Rightarrow\text{DC}=\text{BD}$ [sides opposite to equal angles are equal] $\text{BD}=\text{DC}$
According to Eulid’s axiom, if equals are subtracted from equals, then remainders are also equal. On subtracting Eq. $(ii)$ from Eq. $(i),$
we get $\angle\text{ABC}-\angle4=\angle\text{ACB}-\angle3$
$\Rightarrow\angle1=\angle2$
Now, in $ABCD$, $\angle1=\angle2$
$\Rightarrow\text{DC}=\text{BD}$ [sides opposite to equal angles are equal] $\text{BD}=\text{DC}$
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