Question

write a similarity statement comparing the three triangles to each diagram.
1.

△prs ~△ ______
△prs~ △ ______
△qps~ △ ______
2.

△bce ~△ ______
△bce ~△ ______
△ecd ~△ ______

Explanation:

Step1: Analyze Triangle PRS and Similar Triangles (Diagram 1)

In the first diagram, we have right triangles. $\triangle PRS$ is a right triangle with right angle at $S$. $\triangle PSQ$ also has a right angle at $S$, and $\triangle PRQ$ is the large triangle. By AA (Angle - Angle) similarity criterion, $\angle PRS=\angle PSQ = 90^{\circ}$ and $\angle P$ is common to $\triangle PRS$ and $\triangle PSQ$, so $\triangle PRS\sim\triangle PSQ$. Also, $\angle PRS=\angle PRQ$ (common angle) and $\angle PSR=\angle PQR = 90^{\circ}$, so $\triangle PRS\sim\triangle PRQ$. And $\triangle QPS\sim\triangle QPR$ (by AA similarity as well, since $\angle Q$ is common and both are right triangles at $S$ and $R$ respectively). Wait, let's correct:

For $\triangle PRS$:
- $\triangle PRS$ and $\triangle PSQ$: $\angle PRS=\angle PSQ = 90^{\circ}$, $\angle RPS=\angle SPQ$ (common angle? Wait, no, $\angle RPS$ and $\angle QPS$: Wait, $\triangle PRS$ (right at $S$), $\triangle PSQ$ (right at $S$), and $\triangle PRQ$ (right at $P$? Wait, no, the diagram has $P$ with a right angle, $S$ with a right angle. So $\angle PRS = 90^{\circ}$, $\angle PSQ=90^{\circ}$, $\angle QPR = 90^{\circ}$? Wait, maybe the triangles are $\triangle PRS$, $\triangle PSQ$, and $\triangle PRQ$.

By AA similarity:
- $\triangle PRS \sim \triangle PSQ$ ( $\angle R = \angle S = 90^{\circ}$, $\angle P$ is common? Wait, $\angle RPS$ and $\angle QPS$: No, $\angle PRS$ and $\angle PSQ$ are right angles, and $\angle RSP=\angle SQP$? Wait, maybe better to use the right - triangle similarity. In a right triangle, if we draw an altitude from the right angle to the hypotenuse, the two smaller triangles are similar to the original triangle and to each other. So in $\triangle PRQ$ (right - angled at $P$), with altitude $PS$ to hypotenuse $RQ$, then $\triangle PRS\sim\triangle QPS\sim\triangle PRQ$.

So for $\triangle PRS\sim$:
- First, $\triangle PRS\sim\triangle PSQ$ (because $\angle PRS=\angle PSQ = 90^{\circ}$, $\angle R=\angle SPQ$? Wait, no, $\angle RPS+\angle QPS = 90^{\circ}$, $\angle RPS+\angle R=90^{\circ}$, so $\angle QPS=\angle R$, and $\angle R=\angle SPQ$? Wait, let's use AA: $\angle PRS=\angle PSQ = 90^{\circ}$, $\angle R=\angle SPQ$ (since $\angle R + \angle RPS=90^{\circ}$ and $\angle SPQ+\angle RPS = 90^{\circ}$), so $\triangle PRS\sim\triangle PSQ$.
- Second, $\triangle PRS\sim\triangle PRQ$ ( $\angle PRS=\angle QPR = 90^{\circ}$, $\angle R=\angle R$ (common angle), so AA similarity).
- And $\triangle QPS\sim\triangle QPR$ ( $\angle QPS=\angle QPR$? No, $\angle Q$ is common, $\angle QSP=\angle QRP = 90^{\circ}$, so AA similarity, so $\triangle QPS\sim\triangle QRP$.

For the second diagram ( $\triangle BCE$, $\triangle ECD$, etc.):
$\triangle BCE$ is right - angled at $C$, $\triangle ECD$ is right - angled at $C$, and $\triangle BED$ (or $\triangle BDE$) is the large triangle. By AA similarity, $\triangle BCE\sim\triangle DCE$ (wait, no, $\triangle BCE$ and $\triangle DCE$: $\angle BCE=\angle DCE = 90^{\circ}$, $\angle BEC=\angle DEC$? No, $\triangle BCE$ and $\triangle BED$: $\angle BCE=\angle BED = 90^{\circ}$, $\angle B$ is common, so $\triangle BCE\sim\triangle BED$. $\triangle BCE$ and $\triangle DCE$: $\angle BCE=\angle DCE = 90^{\circ}$, $\angle B=\angle CDE$ (since $\angle B+\angle BEC = 90^{\circ}$ and $\angle CDE+\angle DEC = 90^{\circ}$, and $\angle BEC=\angle DEC$? Wait, maybe the correct similar triangles:

For $\triangle BCE$ (right at $C$):
- $\triangle BCE\sim\triangle DEC$ ( $\angle BCE=\angle DEC = 90^{\circ}$, $\angle B=\angle CDE$)
- $\triangle BCE\sim\triangle BDE$ ( $\angle BCE=\angle BED = 90^{\circ}$, $\angle B$ is common)
- $\triangle ECD\sim\triangle BED$ ( $\angle ECD=\angle BED = 90^{\circ}$, $\angle D$ is common)

Now, let's correct the first diagram:

1. In the first diagram, $\triangle PRS$ (right at $S$), $\triangle PSQ$ (right at $S$), $\triangle PRQ$ (right at $P$). By the geometric mean theorem (right - triangle similarity), when we have a right triangle and an altitude to the hypotenuse, the two smaller triangles are similar to the original triangle and to each other. So:

- $\triangle PRS\sim\triangle PSQ$ (AA: $\angle R=\angle S = 90^{\circ}$, $\angle RPS=\angle SPQ$? Wait, $\angle RPS+\angle QPS = 90^{\circ}$, $\angle RPS+\angle R = 90^{\circ}$, so $\angle QPS=\angle R$, and $\angle R=\angle SPQ$? Maybe better to say $\triangle PRS\sim\triangle QPS$ (no, $\triangle PRS$ and $\triangle QPS$: $\angle R=\angle S = 90^{\circ}$, $\angle P$ is common? Wait, $\triangle PRS$: vertices $P, R, S$; $\triangle PSQ$: vertices $P, S, Q$; $\triangle PRQ$: vertices $P, R, Q$. So $\triangle PRS\sim\triangle PSQ$ ( $\angle PRS=\angle PSQ = 90^{\circ}$, $\angle RPS=\angle QPS$? No, $\angle RPS$ and $\angle QPS$ are angles at $P$, so $\angle RPS+\angle QPS=\angle RPQ$. Wait, maybe the correct similarity statements:

$\triangle PRS\sim\triangle PSQ$ (AA: $\angle PRS=\angle PSQ = 90^{\circ}$, $\angle R=\angle SPQ$ (since $\angle R + \angle RPS=90^{\circ}$ and $\angle SPQ+\angle RPS = 90^{\circ}$))

$\triangle PRS\sim\triangle PRQ$ (AA: $\angle PRS=\angle QPR = 90^{\circ}$, $\angle R=\angle R$ (common angle))

$\triangle QPS\sim\triangle QPR$ (AA: $\angle QPS=\angle QPR = 90^{\circ}$? No, $\angle Q$ is common, $\angle QSP=\angle QRP = 90^{\circ}$)

For the second diagram, $\triangle BCE$ (right at $C$):

- $\triangle BCE\sim\triangle DCE$ (AA: $\angle BCE=\angle DCE = 90^{\circ}$, $\angle B=\angle CDE$)
- $\triangle BCE\sim\triangle BDE$ (AA: $\angle BCE=\angle BED = 90^{\circ}$, $\angle B$ is common)
- $\triangle ECD\sim\triangle BED$ (AA: $\angle ECD=\angle BED = 90^{\circ}$, $\angle D$ is common)

Now, let's fill in the blanks:

1. $\triangle PRS\sim\triangle PSQ$ (first blank)
$\triangle PRS\sim\triangle PRQ$ (second blank)
$\triangle QPS\sim\triangle QPR$ (third blank)

2. $\triangle BCE\sim\triangle DCE$ (first blank)
$\triangle BCE\sim\triangle BDE$ (second blank)
$\triangle ECD\sim\triangle BED$ (third blank)

Wait, maybe the correct similar triangles for the first diagram are:

In $\triangle PRQ$ (right - angled at $P$), with altitude $PS$ to hypotenuse $RQ$, then:

- $\triangle PRS\sim\triangle QPS$ (AA: $\angle PRS=\angle QPS = 90^{\circ}$, $\angle R=\angle QSP$)
- $\triangle PRS\sim\triangle PRQ$ (AA: $\angle PRS=\angle QPR = 90^{\circ}$, $\angle R=\angle R$)
- $\triangle QPS\sim\triangle QPR$ (AA: $\angle QPS=\angle QPR = 90^{\circ}$, $\angle Q=\angle Q$)

For the second diagram, in $\triangle BDE$ (right - angled at $E$), with altitude $EC$ to hypotenuse $BD$, then:

- $\triangle BCE\sim\triangle DCE$ (AA: $\angle BCE=\angle DCE = 90^{\circ}$, $\angle B=\angle CDE$)
- $\triangle BCE\sim\triangle BDE$ (AA: $\angle BCE=\angle BED = 90^{\circ}$, $\angle B=\angle B$)
- $\triangle ECD\sim\triangle BED$ (AA: $\angle ECD=\angle BED = 90^{\circ}$, $\angle D=\angle D$)

So the correct similarity statements:

1.
$\triangle PRS\sim\triangle PSQ$
$\triangle PRS\sim\triangle PRQ$
$\triangle QPS\sim\triangle QPR$

2.
$\triangle BCE\sim\triangle DCE$
$\triangle BCE\sim\triangle BDE$
$\triangle ECD\sim\triangle BED$

Answer:

1. $\triangle PRS\sim\triangle PSQ$; $\triangle PRS\sim\triangle PRQ$; $\triangle QPS\sim\triangle QPR$
2. $\triangle BCE\sim\triangle DCE$; $\triangle BCE\sim\triangle BDE$; $\triangle ECD\sim\triangle BED$