Question

in the figure below, there are three right triangles. complete the following. (a)write a similiarity statement relating the three right triangles. \\(\triangle wyx \sim \triangle \square \sim \triangle \square\\) (b)complete each proportion. \\(\dfrac{zx}{yx} = \dfrac{yx}{\square}\\) \\(\dfrac{wz}{yz} = \dfrac{\square}{zx}\\)

Explanation:

Step1: Analyze Similar Triangles

In a right triangle, when an altitude is drawn to the hypotenuse, the two smaller triangles are similar to the original triangle and to each other. Here, $\triangle WYX$ is the original right triangle, $\triangle XZY$ and $\triangle WZY$ (wait, actually, the triangles are $\triangle XZY$, $\triangle WYZ$, and $\triangle WYX$). Wait, looking at the figure, $\angle XZY = 90^\circ$, $\angle WYZ = 90^\circ$? No, the altitude from $Y$ to $XW$ is $YZ$ (wait, the labels: $X$, $Z$, $W$ on the base, $Y$ at the top. So $\triangle WYX$ is right-angled at $Y$? Wait, no, the right angle is at $Z$ (since $YZ$ is perpendicular to $XW$). So $\triangle XZY$ is right-angled at $Z$, $\triangle WZY$ is right-angled at $Z$, and $\triangle WYX$ is right-angled at $Y$? Wait, no, the original triangle: $\angle XYW$ is right? Wait, no, the figure shows $YZ \perp XW$, so $\angle XZY = \angle WZY = 90^\circ$, and $\angle XYW$ is the right angle? Wait, maybe I mislabel. Let's recall the geometric mean theorem (altitude-on-hypotenuse theorem): In a right triangle, the altitude to the hypotenuse forms two smaller right triangles that are similar to the original triangle and to each other. So if $\triangle WYX$ is right-angled at $Y$, and $YZ \perp XW$, then $\triangle XZY \sim \triangle WYX \sim \triangle WYZ$. Wait, no, the labels: $X$, $Z$, $W$ on the base, $Y$ above. So $\triangle WYX$: vertices $W$, $Y$, $X$. $\triangle XZY$: $X$, $Z$, $Y$. $\triangle WYZ$: $W$, $Y$, $Z$. So by AA similarity: $\angle X$ is common to $\triangle WYX$ and $\triangle XZY$, and both are right-angled, so $\triangle WYX \sim \triangle XZY$. Similarly, $\angle W$ is common to $\triangle WYX$ and $\triangle WYZ$, both right-angled, so $\triangle WYX \sim \triangle WYZ$. Also, $\triangle XZY \sim \triangle WYZ$ by AA (both right-angled, and $\angle XYZ = \angle W$ or something). So for part (a), $\triangle WYX \sim \triangle XZY \sim \triangle WYZ$. Wait, but the first blank is after $\triangle WYX \sim \triangle \square \sim \triangle \square$. So the two smaller triangles are $\triangle XZY$ (or $\triangle YZX$) and $\triangle WYZ$ (or $\triangle YZW$). Wait, maybe the labels are $\triangle YZX$ and $\triangle WYZ$. Let's check the notation: $\triangle WYX$: vertices $W$, $Y$, $X$. So the first similar triangle should be $\triangle YZX$ (since $\angle X$ is common, right angle at $Z$ and $Y$? Wait, no, $\triangle WYX$: right angle at $Y$? Wait, no, the right angle is at $Z$ (because $YZ \perp XW$). So $\triangle WYX$ is right-angled at $Y$? No, that can't be. Wait, maybe the original triangle is right-angled at $Y$, and $YZ$ is the altitude to hypotenuse $XW$. Then by the altitude-on-hypotenuse theorem, $\triangle WYX \sim \triangle YZX \sim \triangle WYZ$. So $\triangle WYX \sim \triangle YZX \sim \triangle WYZ$. So the first blank is $\triangle YZX$ (or $\triangle XZY$) and the second is $\triangle WYZ$ (or $\triangle YZW$).

Step2: Part (a) Similarity Statement

So $\triangle WYX \sim \triangle YZX \sim \triangle WYZ$. Wait, but the notation: $\triangle WYX$: $W$, $Y$, $X$. $\triangle YZX$: $Y$, $Z$, $X$. $\triangle WYZ$: $W$, $Y$, $Z$. So the order of vertices matters for similarity (corresponding angles). So $\angle W$ corresponds to $\angle Y$? Wait, no, let's do AA: $\triangle WYX$ and $\triangle YZX$: $\angle X$ is common, $\angle WYX = \angle YZX = 90^\circ$? Wait, no, $\angle WYX$ is the right angle? If $YZ$ is the altitude, then $\angle YZX = 90^\circ$, and $\angle WYX$ is also $90^\circ$? Then $\angle X$ is common, so AA similarity: $\triangle WYX \sim \triangle YZX$. Similarly, $\triangle WYX$ and $\triangle WYZ$: $\angle W$ is common, $\angle WYX = \angle WYZ = 90^\circ$, so AA similarity: $\triangle WYX \sim \triangle WYZ$. Also, $\triangle YZX \sim \triangle WYZ$ by AA (both right-angled, and $\angle XYZ = \angle W$). So the similarity statement is $\triangle WYX \sim \triangle YZX \sim \triangle WYZ$.

Step3: Part (b) Proportions

First proportion: $\frac{ZX}{YX} = \frac{YX}{WX}$. Wait, no, the geometric mean theorem: in a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which it divides the hypotenuse. Also, each leg of the original triangle is the geometric mean of the hypotenuse and the adjacent segment. So for leg $YX$: $YX^2 = ZX \times WX$? Wait, no, $ZX$ is the segment adjacent to $X$, $WX$ is the hypotenuse. Wait, the proportion $\frac{ZX}{YX} = \frac{YX}{WX}$? Wait, the first proportion is $\frac{ZX}{YX} = \frac{YX}{\square}$. By the geometric mean theorem, $YX^2 = ZX \times WX$? No, wait, $ZX$ is one segment, $WX$ is the hypotenuse? Wait, no, the hypotenuse of $\triangle WYX$ is $WX$, and the segments are $ZX$ and $WZ$. So $YX$ is a leg of $\triangle WYX$, and it's also the hypotenuse of $\triangle YZX$. So in $\triangle YZX$ and $\triangle WYX$, the ratio of corresponding sides: $ZX$ (leg of $\triangle YZX$) over $YX$ (leg of $\triangle WYX$) equals $YX$ (hypotenuse of $\triangle YZX$) over $WX$ (hypotenuse of $\triangle WYX$). So $\frac{ZX}{YX} = \frac{YX}{WX}$.

Second proportion: $\frac{WZ}{YZ} = \frac{YZ}{ZX}$. Wait, no, the proportion is $\frac{WZ}{YZ} = \frac{\square}{ZX}$. Wait, the geometric mean theorem also states that $YZ^2 = ZX \times WZ$, so $\frac{WZ}{YZ} = \frac{YZ}{ZX}$. Wait, but the given proportion is $\frac{WZ}{YZ} = \frac{\square}{ZX}$. So cross-multiplying, $WZ \times ZX = YZ \times \square$, but from $YZ^2 = ZX \times WZ$, we have $WZ \times ZX = YZ^2$, so $\frac{WZ}{YZ} = \frac{YZ}{ZX}$. Wait, no, the proportion is $\frac{WZ}{YZ} = \frac{\square}{ZX}$. So solving for $\square$, $\square = \frac{WZ \times ZX}{YZ}$, but from $YZ^2 = ZX \times WZ$, so $\square = YZ$. Wait, no, let's check the proportion: $\frac{WZ}{YZ} = \frac{YZ}{ZX}$? No, that would be $YZ^2 = WZ \times ZX$. Wait, maybe I mixed up. Wait, the segments: $ZX$ and $WZ$ are the two parts of the hypotenuse $XW$, with $ZX$ adjacent to $X$, $WZ$ adjacent to $W$. The altitude is $YZ$. Then:

- $YX^2 = ZX \times XW$ (wait, $XW$ is the hypotenuse, so $XW = ZX + WZ$). Wait, no, the correct geometric mean formulas are:

- Leg (adjacent to $X$): $YX^2 = ZX \times XW$? No, no, the hypotenuse is $XW$, and the segments are $ZX$ (near $X$) and $WZ$ (near $W$). So:

- $YX^2 = ZX \times XW$? No, that's not right. Wait, the altitude-on-hypotenuse theorem: In a right triangle, the square of a leg is equal to the product of the hypotenuse and the adjacent segment. So if $\triangle WYX$ is right-angled at $Y$, with hypotenuse $XW$, and altitude $YZ$ to $XW$, then:

- $YX^2 = ZX \times XW$? No, $XW$ is the hypotenuse, so $XW = ZX + WZ$. Wait, no, the adjacent segment to $YX$ is $ZX$ (since $YX$ is the leg from $Y$ to $X$, and the segment from $X$ to $Z$ is adjacent). Wait, actually, the correct formula is:

- $YX^2 = ZX \times XW$ (no, that's not, the correct is $YX^2 = ZX \times XW$? Wait, no, let's take an example. Suppose $ZX = 4$, $WZ = 9$, then $YZ = 6$ (since $6^2 = 4 \times 9$), and $YX = \sqrt{4 \times (4 + 9)} = \sqrt{52}$? No, wait, $YX^2 = ZX \times XW$ where $XW = ZX + WZ = 13$, so $YX^2 = 4 \times 13 = 52$, $YW^2 = 9 \times 13 = 117$, and $YZ^2 = 4 \times 9 = 36$, which matches $6^2$. So yes, $YX^2 = ZX \times XW$, $YW^2 = WZ \times XW$, $YZ^2 = ZX \times WZ$.

So for the first proportion: $\frac{ZX}{YX} = \frac{YX}{XW}$. Wait, the given proportion is $\frac{ZX}{YX} = \frac{YX}{\square}$. So $\square$ is $XW$ (or $WX$).

For the second proportion: $\frac{WZ}{YZ} = \frac{YZ}{ZX}$. Wait, the given proportion is $\frac{WZ}{YZ} = \frac{\square}{ZX}$. So cross-multiplying, $WZ \times ZX = YZ \times \square$, but from $YZ^2 = ZX \times WZ$, we have $WZ \times ZX = YZ^2$, so $\square = YZ$. Wait, no, the proportion is $\frac{WZ}{YZ} = \frac{YZ}{ZX}$? No, that would be $YZ^2 = WZ \times ZX$, which is correct. So the missing term is $YZ$.

Answer:

(a) $\triangle WYX \sim \triangle YZX \sim \triangle WYZ$ (or $\triangle XZY$ and $\triangle WYZ$ depending on notation, but likely $\triangle YZX$ and $\triangle WYZ$)
(b) First proportion: $\frac{ZX}{YX} = \frac{YX}{WX}$ (so the blank is $WX$), second proportion: $\frac{WZ}{YZ} = \frac{YZ}{ZX}$ (so the blank is $YZ$)

Wait, but let's confirm the similarity statement again. The three triangles are $\triangle WYX$ (right-angled at $Y$), $\triangle XZY$ (right-angled at $Z$), and $\triangle WZY$ (right-angled at $Z$)? No, $\triangle WZY$ is right-angled at $Z$, and $\triangle XZY$ is right-angled at $Z$. Wait, no, the altitude from $Y$ to $XW$ is $YZ$, so $\angle YZX = \angle YZW = 90^\circ$, and $\angle XYW = 90^\circ$. So by AA similarity:

• $\triangle WYX$ and $\triangle XZY$: $\angle X$ is common, $\angle XYW = \angle XZY = 90^\circ$ ⇒ similar.

• $\triangle WYX$ and $\triangle WZY$: $\angle W$ is common, $\angle XYW = \angle WZY = 90^\circ$ ⇒ similar.

• $\triangle XZY$ and $\triangle WZY$: $\angle XZY = \angle WZY = 90^\circ$, $\angle XYZ = \angle W$ (since $\triangle XZY \sim \triangle WYX$, so $\angle XYZ = \angle W$) ⇒ similar.

So the similarity statement is $\triangle WYX \sim \triangle XZY \sim \triangle WZY$? Wait, maybe the labels are $Z$ between $X$ and $W$, $Y$ above. So $\triangle XZY$ (X, Z, Y), $\triangle WZY$ (W, Z, Y), and $\triangle WYX$ (W, Y, X). So $\triangle WYX \sim \triangle XZY$ (corresponding vertices: W→X, Y→Z, X→Y? No, that's not. Wait, maybe the correct order is $\triangle WYX \sim \triangle YZX \sim \triangle WYZ$. I think the key is that the three triangles are similar, so the answer for (a) is $\triangle YZX$ (or $\triangle XZY$) and $\triangle WYZ$ (or $\triangle YZW$).

For part (b), the first proportion: by the geometric mean theorem, $ZX/YX = YX/WX$ (since $YX$ is a leg, $ZX$ is the adjacent segment, $WX$ is the hypotenuse). The second proportion: $WZ/YZ = YZ/ZX$ (since $YZ$ is the altitude, $WZ$ and $ZX$ are the segments, so $YZ^2 = WZ \times ZX$ ⇒ $WZ/YZ = YZ/ZX$).

So final answers:

(a) $\triangle WYX \sim \triangle YZX \sim \triangle WYZ$ (or $\triangle XZY$ and $\triangle WYZ$)

(b) First blank: $WX$; second blank: $YZ$