Exercise 3 Page 387

Rhombuses

Let's consider a rhombus PQRS and let T be the intersection point of the two diagonals.

Let's list the congruent parts between △ RST and △ RQT. ccc RS ≅ RQ & & Side ST ≅ QT & & Side TR ≅ TR & & Side By the Side-Side-Side (SSS) Congruence Theorem, we have that △ RST ≅ △ RQT. Consequently, ∠ STR ≅ ∠ RTQ.

Since ∠ STR and ∠ RTQ form a linear pair and they are congruent, we have that they are right angles. m ∠ STR + m ∠ RTQ = 180^(∘) ⇓ m ∠ STR = m ∠ RTQ = 90^(∘) From the above, we can write the following conclusion.

Additionally, from the triangle congruence, we have that ∠ TRS ≅ ∠ TRQ, which means that PR bisects ∠ QRS.

By repeating this process with the other three pairs of triangles, we will get the same conclusion. That is, each diagonal bisects two opposite angles of the rhombus.

Squares

Finally, let's consider a square WXYZ.

Notice that a square is both a rectangle and a rhombus at the same time. Then, from the two cases above, we can write the following statement.