In a triangle, prove the perpendicular from the vertex opposite the side that is not less than either of the remaining sides falls within the triangle.
it is clearly true for all acute triangles.
In a right or obtuse triangle, the longest side is opposite the largest angle.
That vertex lies above the longest side, so the perpendicular meets the side within the triangle.
To prove that the perpendicular from the vertex opposite the side that is not less than either of the remaining sides falls within the triangle, we can use the concept of inequalities and the properties of triangles.
Let's consider a triangle with vertices A, B, and C. We want to prove that the perpendicular from vertex C, opposite side c, falls within the triangle. Here, a, b, and c are the lengths of the respective sides.
To begin, we'll assume that side c is not less than either of the remaining sides, a and b. In other words, we assume c ≥ a and c ≥ b.
Now, let's draw the perpendicular from vertex C to side AB and label the point of intersection as D. We want to show that point D lies within triangle ABC.
To do this, we'll use contradiction. We'll assume that point D is located outside the triangle ABC and falls either on side AB or outside of it. We'll consider both cases separately.
Case 1: Point D lies on side AB
In this case, D lies on the line segment AB but not within the triangle. Without loss of generality, let's assume that D lies to the right of the midpoint of AB. Let M be the midpoint of AB.
Since D lies outside the triangle, triangle ACD will be larger than triangle ABC. Therefore, the side AC will be longer than side AB, contradicting the assumption that c ≥ a.
Case 2: Point D lies outside the line segment AB
In this case, D lies outside the line segment AB. Without loss of generality, we'll assume that D is to the right of side AB (the reasoning would be similar if D is to the left).
Since D lies outside the triangle, triangle BCD will be larger than triangle ABC. Therefore, the side BC will be longer than side AB, contradicting the assumption that c ≥ b.
Since both cases lead to a contradiction, our assumption that point D lies outside the triangle ABC must be incorrect. Therefore, the perpendicular from vertex C must fall within the triangle ABC when c ≥ a and c ≥ b.
This completes the proof.