Question

ABC is a triangle and D is the md-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.

Answer 1

Hint: In this problem, we need to apply the congruence configuration to prove the \[\Delta ABC\] as isosceles triangle. The sides of a triangle are equal if corresponding opposite angles are equal.

Complete step by step answer:
In \[\Delta DFC\] and \[\Delta DEB\],
\[
  BD = DC\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {D\,\,{\text{is mid point}}} \right) \\
  DE = DF\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Given}}} \right) \\
\]
Thus, the \[\Delta DFC\] and \[\Delta DEB\] are congruent.
\[
  \,\,\,\,\,\angle DCF = \angle DBE \\
   \Rightarrow \angle BCA = \angle CBA \\
  \therefore AB = AC \\
\]

Thus, the \[\Delta ABC\] is an isosceles triangle.

Note: Two triangles are said to be congruent if the sides and angles of the triangles are exactly the same. If all the sides of the triangles are the same, it is also termed a congruent triangle. Two triangles with same angles are said to be congruent if sides of the triangle are of same dimensions.