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Answer :

In \(\triangle{ABC}\), we have,

AB = AC ...(\(\because \) \(\triangle{ABC}\) is an isosceles triangle)

\(\therefore \) \(\angle{ABC}\) = \(\angle{ACB}\) ...(i)(\(\because \) angles opposite to equal sides are equal)

Similarly, in \(\triangle{DBC}\), we have,

BD = CD ...(\(\because \) \(\triangle{DBC}\) too, is an isosceles triangle)

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

Now, On adding, Equations (i) and (ii), we get,

\(\Rightarrow \angle{ABC} + \angle{DBC} = \angle{ACB} + \angle{DCB} \)

\(\therefore \) \(\angle{ABD}\) = \(\angle{ACD}\)

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.
- \(\triangle{ABC}\) 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 \(\angle{C}\).
- Show that the angles of an equilateral triangle are \(60^\circ\) each.

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