We are given a triangle $LMN$ with $LP$ as an altitude to side $NM$. We know that $NP=9$ and $LM = 6$. We are asked to find two triangles that are similar to triangle $LMN$ and to find the length of $PM$.

1. Problem Description

We are given a triangle $LMN$ with $LP$ as an altitude to side $NM$. We know that $NP=9$ and $LM = 6$.

We are asked to find two triangles that are similar to triangle $LMN$ and to find the length of $PM$.

2. Solution Steps

(i) Find similar triangles:

Since $LP$ is an altitude to side $NM$, $\angle LPN = 90^{\circ}$ and $\angle LPM = 90^{\circ}$.

Also, $\angle LLN = 90^{\circ}$.

Consider $\triangle LMN$. $\angle LLN$ is a right angle.

Consider $\triangle LPN$. $\angle LPN$ is a right angle.

Consider $\triangle LPM$. $\angle LPM$ is a right angle.

Now, we compare $\triangle LMN$ and $\triangle LPN$.

$\angle LMN = \angle LMN$

$\angle LLN = \angle LPN = 90^{\circ}$

So, $\triangle LMN \sim \triangle LPN$ by AA similarity.

Next, we compare $\triangle LMN$ and $\triangle LPM$.

$\angle LMN = \angle LML$

$\angle LLN = \angle LPM = 90^{\circ}$

So, $\triangle LMN \sim \triangle LPM$ by AA similarity.

Thus, $\triangle LPN \sim \triangle LPM$.

Two triangles similar to $\triangle LMN$ are $\triangle LPN$ and $\triangle LPM$.

(ii) Find $PM$:

Since $\triangle LMN \sim \triangle LPM$, we have:

$\frac{LM}{PM} = \frac{MN}{LM}$

$LM^2 = PM \cdot MN$

$6^2 = PM \cdot (NP + PM)$

$36 = PM \cdot (9 + PM)$

$36 = 9PM + PM^2$

$PM^2 + 9PM - 36 = 0$

$(PM + 12)(PM - 3) = 0$

$PM = -12$ or $PM = 3$.

Since the length cannot be negative, $PM = 3$.

3. Final Answer

The two triangles similar to $\triangle LMN$ are $\triangle LPN$ and $\triangle LPM$.

$PM = 3$