Question

$\overrightarrow{cr}$ and $\overrightarrow{ds}$ are perpendiculars dropped from $\overleftrightarrow{ab}$ to $\overleftrightarrow{pq}$, and $\overleftrightarrow{ab}$ is perpendicular to $\overrightarrow{cr}$ and $\overrightarrow{ds}$. if $cr = ds$, which statement must be true?
a. $m\angle rcd=m\angle sdb + 2$
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 \(CR\perp PQ\) and \(AB\) is a straight line, \(\angle RCD + \angle ACD= 180^{\circ}\) (linear - pair of angles).

Step2: Consider the properties of perpendiculars

We know that \(CR = DS\), \(CR\perp PQ\), \(DS\perp PQ\), and \(AB\parallel PQ\) (because \(AB\perp CR\) and \(CR\perp PQ\)). The quadrilateral \(CRSD\) is a rectangle (since \(CR\parallel DS\), \(CR = DS\), and \(CR\perp RS\), \(DS\perp RS\)). So \(CD\) is a straight line segment.
If we assume \(\angle RCD=x\) and \(\angle ACD = y\), then \(x + y=180^{\circ}\). But if we consider the fact that \(CD\) is a straight line segment and there is no other angular relation given except the linear - pair. However, if we assume \(CD\) is a bisector (in the case of a rectangle - like formation where \(CR = DS\) and the perpendiculars), we note that \(\angle RCD\) and \(\angle ACD\) are supplementary. But if we consider the fact that \(CD\) is a straight line and there is no angle - addition or - subtraction other than the linear - pair. Wait, actually, since \(CR\perp AB\) (because \(AB\) is perpendicular to \(CR\) and \(DS\)), \(\angle RCD+\angle ACD = 180^{\circ}\) is wrong. Wait, no, \(AB\) is a straight line. \(\angle RCD\) and \(\angle ACD\) form a linear pair. But if we consider the fact that \(CR = DS\), \(CR\perp PQ\), \(DS\perp PQ\), and \(AB\) is a straight line. The measure of \(\angle RCD = 90^{\circ}\) (because \(CR\perp PQ\) and \(AB\parallel PQ\) (since \(AB\perp CR\) and \(CR\perp PQ\)) and \(\angle ACD = 90^{\circ}\)

Answer:

B. \(m\angle RCD=m\angle ACD\)