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Show that these altitudes are equal.

Answer :

In \(\triangle{ABE}\) and \(\triangle{ACF}\) , we have,

\(\angle{AEB}\) = \(\angle{AFC}\) ...(BE and CF are perpendiculars drawn to sides AC and AB respectively)

Also, \(\angle{A}\) is common angle.

and AB = AC ...(Given)

Therefore, \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) ...(By AAS congruency test)

Therefore, AB = AC ...(By CPCT)

Hence, BE = CF ...(By CPCT)

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

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