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Show that \(\angle{BCD}\) is a right angle.

Answer :

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

AB = AC ...(i)(Given)

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

Also, AB = AD ...(iii)(Given)

From (i) and (iii), AC = AD

Now, in \(\triangle{ADC}\) , we have,

AD = AC ...(proved earlier)

\(\angle{ACD}\) = \(\angle{ADC}\) ...(\(\because \) angles opposite to equal sides are equal)

Also, \(\angle{ACD}\) = \(\angle{BDC}\) ...(iv)(\(\because \angle{ADC} = \angle{BDC}\))

On adding Equations (ii) and (iv), we get,

\(\Rightarrow \angle{ACB} + \angle{ACD} = \angle{ABC} + \angle{BDC} \)

\(\Rightarrow \) \(\angle{BCD}\) = \(\angle{ABC}\) + \(\angle{BDC}\)

Adding \(\angle{BCD}\) on both sides, we have,

\( \Rightarrow \angle{BCD} + \angle{BCD} = \angle{ABC} + \angle{BDC} + \angle{BCD} \)

\(\Rightarrow \) 2 \(\angle{BCD}\) = \(180^\circ\) ...(\(\because \) sum of all angles of a triangle is \(180^\circ\))

Therefore, \(\angle{BCD}\) = \(90^\circ\)

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 \(\angle{ABD}\) = \(\angle{ACD}\).
- 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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