The problem provides a right triangle $\triangle LMN$ with $\angle L = 90^\circ$. A perpendicular line segment $LP$ is drawn from $L$ to $NM$. Given $NP = 9$ and $LM = 6$, the problem asks to: (i) name two triangles similar to $\triangle LMN$, (ii) find the length of $PM$.

1. Problem Description

The problem provides a right triangle $\triangle LMN$ with $\angle L = 90^\circ$. A perpendicular line segment $LP$ is drawn from $L$ to $NM$. Given $NP = 9$ and $LM = 6$, the problem asks to:

(i) name two triangles similar to $\triangle LMN$,

(ii) find the length of $PM$.

2. Solution Steps

(i) Similar triangles:

Since $\angle L = 90^\circ$ and $LP \perp NM$, we have three right triangles: $\triangle LMN$, $\triangle LPN$, and $\triangle LMP$.

$\triangle LMN \sim \triangle LPN$ because they share $\angle N$ and both have a right angle ($\angle L$ and $\angle LPN$ respectively).

$\triangle LMN \sim \triangle LMP$ because they share $\angle M$ and both have a right angle ($\angle L$ and $\angle LPM$ respectively).

Also, $\triangle LPN \sim \triangle LMP$ because both are similar to $\triangle LMN$.

(ii) Finding $PM$:

Since $\triangle LMN \sim \triangle LMP$, the ratios of corresponding sides are equal:

$\frac{LM}{NM} = \frac{PM}{LM}$.

We have $LM = 6$ and $NM = NP + PM = 9 + PM$. Let $PM = x$.

Then $\frac{6}{9+x} = \frac{x}{6}$.

Cross-multiplying, we have $36 = x(9+x)$, which simplifies to $x^2 + 9x - 36 = 0$.

Factoring the quadratic, we have $(x+12)(x-3) = 0$.

Therefore, $x = -12$ or $x = 3$. Since length cannot be negative, $x = 3$.

Thus, $PM = 3$.

Alternatively, since $\triangle LPN \sim \triangle LMP$, we have

$\frac{NP}{LP} = \frac{LP}{PM}$, which gives $LP^2 = NP \cdot PM$.

Also, since $\triangle LMN \sim \triangle LPN$, we have $\frac{NM}{LM} = \frac{LM}{PM}$.

Let $PM = x$. Then $NM = 9+x$. From $\frac{LP}{NP} = \frac{MP}{LP}$, we get $LP^2 = NP \cdot MP = 9x$.

Also, since $\triangle LMN$ is a right triangle, we have $LN^2 + LM^2 = NM^2$. Also $NM = 9+x$.

Since $\triangle LMP$ is a right triangle, we have $LM^2 = LP^2 + PM^2$.

Then $6^2 = LP^2 + x^2$ or $36 = LP^2 + x^2$, so $LP^2 = 36 - x^2$.

From $LP^2 = 9x$, we get $36 - x^2 = 9x$. Then $x^2 + 9x - 36 = 0$.

Factoring, we get $(x+12)(x-3) = 0$.

Since $x$ must be positive, $x = 3$.

Therefore, $PM = 3$.

3. Final Answer

(i) $\triangle LPN$ and $\triangle LMP$

(ii) $PM = 3$