Chapter 8.1, Problem 40WE

To determine

To prove: If an altitude is drawn from the vertex opposite to the hypotenuse of a right triangle, then two triangles so formed are similar to the original triangle and also to each other.

Explanation of Solution

Given information: $\Delta ABC$ is a right angled triangle, $\angle B$ is of $90°$ , and BX is an altitude.

Formula used: The sum of internal angles of a triangle is always 180°.

Proof: Consider the following figure:

  

Sum of internal angles of a triangle is 180°.

So,

  $\begin{array}{l}\angle p+\angle q+90°=180°,\\ \Rightarrow \angle p+\angle q=180°-90°=90°.\end{array}$

Angle p and q form a complementary pair, that is, one acute angle in right angle triangle will always be compliment of the other acute angle.

Now in (AXB and (BXC,

  $\begin{array}{l}\angle AXB\sim \angle BXC(Sincebotharerightangles)\\ \angle XBA\sim \angle XCB(Since\angle q=\angle XCB)\\ \angle XAB\sim \angle XBC(Since\angle p=\angle XAB)\end{array}$

Using AAA, the triangle $\Delta AXB\sim \Delta BXC$ .

In (AXB, (BXC, and (ABC all corresponding angles are congruent.

Thus, by AAA criterion for similar triangles, $\Delta AXB\sim \Delta BXC\sim \Delta ABC$ .