Chapter 15.2, Problem 15PSB

To determine

To prove: that an altitude of an acute triangle is shorter than either side that is not the base.

Answer to Problem 15PSB

An altitude of an acute triangle is shorter than either side that is not the base.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Acute }\Delta ABC,\\ \text{with base }\overline{BC},\\ \text{alt }\overline{AX}\end{array}$

Calculation:

A triangle $\text{Acute }\Delta ABC,\text{with base }\overline{BC},\text{alt }\overline{AX}$ . Thus,

  

Now, it is to prove $\overline{AX}<\overline{AB},\overline{AX}<\overline{AC}$ .

  $\begin{array}{l}\overline{AX}\perp \overline{BC}\{alt.\text{ form the vertex of a triangle is }\perp \text{ to the opp}\text{. site}\}\\ \angle AXB\text{ is a right}\angle \{\perp linesformrt\angle s\}\\ \angle B<{90}^{\circ }\{all\text{ angles of the acute triangles are equate}}\\ \overline{AX}<\overline{AB}\\ \angle AXB\text{ is a right }\angle s\\ \angle C<{90}^{\circ }\\ \overline{AX}<\overline{AC}\\ \therefore \overline{AX}<\overline{AB},\overline{AX}<\overline{AC}\end{array}$

Thus, an altitude of an acute triangle is shorter than either side that is not the base.