In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $\overline{PR}$. If $PQ = 4QR$,prove that $PD = 16RD$.
(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
$2$. By the property of similarity in right-angled triangles,$\Delta PDQ \sim \Delta QDR \sim \Delta PQR$.
$3$. From $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$4$. From the ratio $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = QD \cdot \frac{PQ}{QR}$.
$5$. From the ratio $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = RD \cdot \frac{PQ}{QR}$.
$6$. Substituting $QD$ in the expression for $PD$: $PD = (RD \cdot \frac{PQ}{QR}) \cdot \frac{PQ}{QR} = RD \cdot (\frac{PQ}{QR})^2$.
$7$. Given $PQ = 4QR$,so $\frac{PQ}{QR} = 4$.
$8$. Therefore,$PD = RD \cdot (4)^2 = 16RD$.
$2$. By the property of similarity in right-angled triangles,$\Delta PDQ \sim \Delta QDR \sim \Delta PQR$.
$3$. From $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$4$. From the ratio $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = QD \cdot \frac{PQ}{QR}$.
$5$. From the ratio $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = RD \cdot \frac{PQ}{QR}$.
$6$. Substituting $QD$ in the expression for $PD$: $PD = (RD \cdot \frac{PQ}{QR}) \cdot \frac{PQ}{QR} = RD \cdot (\frac{PQ}{QR})^2$.
$7$. Given $PQ = 4QR$,so $\frac{PQ}{QR} = 4$.
$8$. Therefore,$PD = RD \cdot (4)^2 = 16RD$.
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For the correspondence $ABC \leftrightarrow XYZ$ being a similarity, which of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$ $1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$ $a. SSS$ condition $2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$ $b. SAS$ condition $3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$ $c. AAA$ condition - $d. \text{None of the conditions apply}$
| Part $I$ | Part $II$ |
|---|---|
| $1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$ | $a. SSS$ condition |
| $2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$ | $b. SAS$ condition |
| $3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$ | $c. AAA$ condition |
| - | $d. \text{None of the conditions apply}$ |
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