Medium
Solve the following question using an appropriate Euclid's axiom:
In the figure,we have $\angle ABC = \angle ACB$ and $\angle 3 = \angle 4$. Show that $\angle 1 = \angle 2$.
In the figure,we have $\angle ABC = \angle ACB$ and $\angle 3 = \angle 4$. Show that $\angle 1 = \angle 2$.
(N/A) We are given:
$\angle ABC = \angle ACB$ ...$(1)$
$\angle 3 = \angle 4$ ...$(2)$
According to Euclid's Axiom $3$,if equals are subtracted from equals,the remainders are equal.
Subtracting equation $(2)$ from equation $(1)$,we get:
$\angle ABC - \angle 4 = \angle ACB - \angle 3$
From the figure,$\angle ABC - \angle 4 = \angle 1$ and $\angle ACB - \angle 3 = \angle 2$.
Therefore,$\angle 1 = \angle 2$.
$\angle ABC = \angle ACB$ ...$(1)$
$\angle 3 = \angle 4$ ...$(2)$
According to Euclid's Axiom $3$,if equals are subtracted from equals,the remainders are equal.
Subtracting equation $(2)$ from equation $(1)$,we get:
$\angle ABC - \angle 4 = \angle ACB - \angle 3$
From the figure,$\angle ABC - \angle 4 = \angle 1$ and $\angle ACB - \angle 3 = \angle 2$.
Therefore,$\angle 1 = \angle 2$.
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