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Answer :

Let \(\triangle{ABC}\) be an equilateral triangle, such that

AB = BC = CA (by property)

Now, we have,

AB = AC

\(\angle{B}\) = \(\angle{C}\) ...(i)(\(\because \) angles opposite to equal sides are equal)

Similarly,

CB = CA

\(\therefore \) \(\angle{A}\) = \(\angle{B}\) ...(ii)(\(\because \) angles opposite to equal sides are equal)

\(\angle{A}\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\) ...(iii)(\(\because \) the sums of all angles of a triangle are \(180^\circ\))

From Equations (i),(ii) and (iii), we have,

\(\angle{A}\) + \(\angle{A}\) + \(\angle{A}\) = \(180^\circ\)

\(\Rightarrow \) 3 \(\angle{A}\) = \(180^\circ\)

\(\Rightarrow \) \(\angle{A}\) = \(60^\circ\)

\(\Rightarrow \angle{A} = \angle{B} = \angle{C} = 60^\circ\)

Hence, proved.

- In an isosceles triangle ABC, with AB = AC, the bisectors of \(\angle{B}\) and \(\angle{C}\) intersect each other at O. Join A to O. Show that, (i) OB = OC (ii) AO bisects \(\angle{A}\)
- In the \(\triangle{ABC}\), AD is the perpendicular bisector of BC (see figure). Show that \(\triangle{ABC}\) is an isosceles triangle in which AB = AC.
- ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure).Show that these altitudes are equal.
- ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that(i) \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) (ii) AB = AC i.e., ABC is an isosceles triangle.
- ABC and DBC are isosceles triangles on the same base BC (see figure). Show that \(\angle{ABD}\) = \(\angle{ACD}\).
- \(\triangle{ABC}\) is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that \(\angle{BCD}\) is a right angle.
- ABC is a right angled triangle in which \(\angle{A}\) = \(90^\circ\) and AB = AC, find \(\angle{B}\) and \(\angle{C}\).

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