Exercise 1 Page 800

We know that the given parallelogram is a rhombus. To help with the process of solving, let's label the point of intersection of the diagonals as E.

Since the parallelogram is a rhombus, each of its diagonals bisects a pair of opposite angles. We can see that the diagonal BD bisects ∠ ABC into ∠ ABE and ∠ EBC. Since m∠ ABC = 70^(∘), we can calculate the measure of ∠ ABE.

m∠ ABE = 70^(∘)2 ⇔ m∠ ABE = 35^(∘) Furthermore, the diagonals of a rhombus are perpendicular. Therefore, we know that ∠ BEA is a right angle. This means that m∠BEA= 90^(∘).

Now, notice that ∠ 1, ∠ ABE, and ∠ BEA are three angles of a triangle. By the Interior Angles Sum Theorem, we can conclude that their measures add to 180. m ∠ 1 + m ∠ ABE + m∠ BEA=180 We already know that m ∠ ABE = 35 and m∠ BEA = 90, so let's substitute these values into our equation to find m ∠ 1.

m ∠ 1 + m ∠ ABE + m∠ BEA=180

SubstituteII

m ∠ ABE= 35, m ∠ BEA= 90

m ∠ 1 + 35 + 90=180

AddTerms

Add terms

m ∠ 1 + 125 =180

SubEqn

LHS-125=RHS-125

m ∠ 1 = 55

The measure of ∠ 1 is 55^(∘), which corresponds to choice B.