Question

$\overline{cr}$ and $\overline{ds}$ are perpendiculars dropped from $\overleftrightarrow{ab}$ to $\overleftrightarrow{pq}$, and $\overleftrightarrow{ab}$ is perpendicular to $\overline{cr}$ and $\overline{ds}$. if $cr = ds$, which statement must be true?

a. $m\angle rcd=m\angle sdb\div2$
b. $m\angle rcd=m\angle acd$
c. $m\angle rcd=m\angle acd\div2$
d. $m\angle rcd=m\angle acd\div3$
e. $m\angle rcd=m\angle acd\times2$

Explanation:

Step1: Analyze the angles

Since \( \overline{CR}\) and \( \overline{DS}\) are perpendiculars to \( \overline{PQ}\), and \( \overline{AB}\) is perpendicular to \( \overline{CR}\) and \( \overline{DS}\), we know that \( \angle ACR = 90^{\circ}\) and \( \angle BDS = 90^{\circ}\).

Step2: Use the property of a straight - line angle

We know that \( \angle ACD+\angle RCD = 180^{\circ}\) (because \(A\), \(C\), \(D\), \(B\) are collinear, so \( \angle ACD\) and \( \angle RCD\) form a linear pair).
Let \(x = m\angle ACD\) and \(y=m\angle RCD\). Then \(x + y=180^{\circ}\).
If we assume \(x = 60^{\circ}\), then \(y = 120^{\circ}\) (for example).
We can also note that \( \angle ACD\) and \( \angle RCD\) are supplementary.
Let's check each option:

• Option A: \(m\angle RCD=m\angle SDB\div2\). There is no information to support a relationship between \( \angle RCD\) and \( \angle SDB\) in this form.
• Option B: \(m\angle RCD=m\angle ACD\). Since \( \angle ACD+\angle RCD = 180^{\circ}\), unless \( \angle ACD=\angle RCD = 90^{\circ}\) (which is not given), this is false.
• Option C: Let \(m\angle ACD=x\) and \(m\angle RCD = y\). If \(y=x + 2\), then \(x+(x + 2)=180\), \(2x=178\), \(x = 89\), \(y = 91\). But in general, from \(x + y=180\), if \(y=x + 2\), it's a special - case solution, not a must - be - true statement.
• Option D: Let \(m\angle ACD=x\) and \(m\angle RCD = y\). If \(y=x + 3\), then \(x+(x + 3)=180\), \(2x=177\), \(x=88.5\), \(y = 91.5\). But in general, from \(x + y=180\), if \(y=x + 3\), it's a special - case solution, not a must - be - true statement.
• Option E: Let \(m\angle ACD=x\) and \(m\angle RCD = y\). If \(y = 2x\), then \(x+2x=180\) (because \(x + y=180\)). Using the equation \(x + y=180\) and substituting \(y = 2x\) gives \(3x=180\), \(x = 60\), \(y = 120\). In general, from \(x + y=180\) and \(y = 2x\), we can solve \(x+\ 2x=180\) (by substitution).

Answer:

E. \(m\angle RCD=m\angle ACD\times2\)