Question

$\overline{yq}$ and $\overline{xp}$ are altitudes to the congruent sides of isosceles triangle $\triangle wxy$.

select all the true statements.

a. $\frac{wq}{xq}=\frac{wy}{xy}$

b. $yq = xp$

c. $m\angle wxy=m\angle wyx$

d. $\frac{wq}{qx}=\frac{wp}{py}$

e. $\overline{pq}$ is parallel to $\overline{xy}$

f. $m\angle pxy=m\angle qyx$

Explanation:

Step1: Analyze option A

Since \(\triangle WXY\) is isosceles, \(\angle WXY=\angle WYX\), but \(\frac{WQ}{XQ}=\frac{WY}{XY}\) is not correct by triangle similarity or other properties.

Step2: Analyze option B

In \(\triangle WXY\), \(WY = WX\) (isosceles triangle), \(\angle WQY=\angle WPX = 90^{\circ}\), \(\angle W=\angle W\). By AAS (Angle - Angle - Side) congruence, \(\triangle WQY\cong\triangle WPX\). So \(YQ = XP\).

Step3: Analyze option C

In an isosceles triangle \(\triangle WXY\) with \(WX = WY\), by the property of isosceles triangles, \(m\angle WXY=m\angle WYX\).

Step4: Analyze option D

Since \(\triangle WQY\cong\triangle WPX\), \(WQ = WP\) and \(QX=PY\). So \(\frac{WQ}{QX}=\frac{WP}{PY}\).

Step5: Analyze option E

\(\triangle WQY\cong\triangle WPX\), so \(\frac{WQ}{WX}=\frac{WP}{WY}\) (because \(WX = WY\)). By the converse of the basic proportionality theorem (Thales' theorem), \(PQ\parallel XY\).

Step6: Analyze option F

\(\triangle WQY\cong\triangle WPX\), \(\angle QXY=\angle PYX\) (from \(\angle WXY=\angle WYX\) and \(\angle WQX=\angle WPY = 90^{\circ}\)). Let \(\angle WXY=\angle WYX=\alpha\), \(\angle QXY=\angle PYX=\beta\). Then \(m\angle PXY=m\angle QYX=\alpha-\beta\).

Answer:

B. \(YQ = XP\), C. \(m\angle WXY=m\angle WYX\), D. \(\frac{WQ}{QX}=\frac{WP}{PY}\), E. \(\overline{PQ}\) is parallel to \(\overline{XY}\), F. \(m\angle PXY = m\angle QYX\)