ABC is a triangle BD & CE are altitudes drawn from the vertex B & C. Prove that ABC is an isoceles triangle

So far, ABC is just any triangle. The addition of the altitudes do not help.

If it is further specified that BD intersects the segment AC at D, and CE intersects segment AB at E, and that BD and CE are equal, then it is possible to prove that triangles EBC and DCB are congruent.
(Use Pythagoras to prove that the third sides are equal, and use SSS for congruence).

Gjfguu

The
Gdjjf
Dhhdzu
Thgf
Gvvk
Ghk

To prove that triangle ABC is an isosceles triangle, we need to show that two sides of the triangle are equal.

Let's begin by looking at the altitudes BD and CE.

First, we notice that altitudes BD and CE are perpendicular to their respective sides AC and AB. This is because an altitude drawn from a vertex (B and C) to the opposite side (AC and AB) is always perpendicular to that side.

Now, let's consider the right-angled triangles ABD and ACE.

In triangle ABD, we have the right angle at D (because BD is perpendicular to AC).

Similarly, in triangle ACE, we have the right angle at E (because CE is perpendicular to AB).

Since the right angles are equal in both triangles, triangle ABD and triangle ACE are similar triangles (by the Angle-Angle similarity criterion).

Now, let's compare the corresponding sides of the similar triangles:

1. AB/AC = BD/CE

We know that BD and CE are the altitudes drawn from vertices B and C respectively. By definition, the length of an altitude is the perpendicular distance from the vertex to the opposite side. Therefore, BD and CE are the heights of triangles ABC. Since both BD and CE are altitudes from the same base (AC), they are equal in length:

2. BD = CE

So, substituting BD = CE in equation 1, we have:

AB/AC = CE/CE

AB/AC = 1

This implies that AB = AC.

Therefore, we can conclude that triangle ABC is an isosceles triangle, as two of its sides (AB and AC) are equal in length.