Geometry practice worksheet with problems on angle measures, triangle congruence, and incenter properties.

Here are the step-by-step solutions for each problem on the worksheet.

Examples 1 and 2: Angle Bisectors and Perpendicular Bisectors

1. Find $m\angle ABE$
* Reasoning: Point $E$ is on the angle bisector of $\angle ABC$ because it is equidistant from the sides (indicated by the tick marks on segments $AE$ and $CE$). This means the ray $BE$ splits the angle into two equal parts.
* Calculation: Since $m\angle EBC = 43^\circ$, then $m\angle ABE$ must also be $43^\circ$.
* Answer: $43^\circ$

2. Find $m\angle YBA$
* Reasoning: Point $A$ is equidistant from the sides of $\angle YBF$ (segments $AY$ and $AF$ are both length 8). Therefore, ray $BA$ is an angle bisector.
* Calculation: The angle bisector divides the angle into two congruent angles. We are given that $m\angle ABF = 47^\circ$. Therefore, $m\angle YBA$ is also $47^\circ$.
* Answer: $47^\circ$

3. Find $MK$
* Reasoning: Point $K$ lies on the perpendicular bisector of segment $MT$ (indicated by the right angles at $M$ and $T$ and the fact that $VK$ is a shared side in the triangles). According to the Perpendicular Bisector Theorem, any point on the perpendicular bisector is equidistant from the endpoints of the segment. So, $MK = TK$.
* Calculation:
$$2x + 1 = 3x - 8$$
Subtract $2x$ from both sides:
$$1 = x - 8$$
Add 8 to both sides:
$$x = 9$$
Now substitute $x$ back into the expression for $MK$:
$$MK = 2(9) + 1 = 18 + 1 = 19$$
* Answer: $19$

4. Find $m\angle EWL$
* Reasoning: Point $L$ is equidistant from the sides of $\angle EWJ$ (segments $LE$ and $LJ$ have tick marks). Thus, ray $WL$ bisects $\angle EWJ$. This means the two smaller angles are equal: $(7x + 5)^\circ = (3x + 21)^\circ$.
* Calculation:
$$7x + 5 = 3x + 21$$
Subtract $3x$ from both sides:
$$4x + 5 = 21$$
Subtract 5 from both sides:
$$4x = 16$$
Divide by 4:
$$x = 4$$
Now find the measure of $\angle EWL$:
$$7(4) + 5 = 28 + 5 = 33$$
* Answer: $33^\circ$

5. Find $IU$
* Reasoning: Ray $RU$ bisects $\angle TRI$ (both angles are $19^\circ$). Also, there are right angles at $T$ and $I$. By the Angle Bisector Theorem, any point on the angle bisector is equidistant from the sides of the angle. Therefore, $TU = IU$.
* Calculation:
$$2x + 5 = 7x$$
Subtract $2x$ from both sides:
$$5 = 5x$$
Divide by 5:
$$x = 1$$
Now find the length of $IU$:
$$IU = 7(1) = 7$$
* Answer: $7$

6. Find $m\angle MYW$
* Reasoning: Point $Y$ is equidistant from the sides of $\angle PYW$ (segments $YP$ and $YW$ have tick marks). Wait, looking closely at the diagram, the tick marks are on segments $MP$ and $MW$? No, the tick marks are on the segments from $Y$ to the lines? Actually, let's look at the angles. The diagram shows ray $YM$ bisecting $\angle PYW$ because $M$ is equidistant from the rays $YP$ and $YW$ (perpendicular distances $MP$ and $MW$ are marked equal). Therefore, the two angles at vertex $Y$ are equal.
* Calculation:
$$4x - 1 = 2x + 5$$
Subtract $2x$ from both sides:
$$2x - 1 = 5$$
Add 1 to both sides:
$$2x = 6$$
Divide by 2:
$$x = 3$$
Now find $m\angle MYW$:
$$2(3) + 5 = 6 + 5 = 11$$
* Answer: $11^\circ$

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Example 3: Incenter Problems

7. Find $m\angle ARU$
* Reasoning: Point $A$ is the incenter of $\triangle PQR$. This means ray $RA$ bisects $\angle PRQ$. We are given that one part of the angle, $\angle KRA$ (or similar, based on the position), is $40^\circ$. Looking at the diagram, the angle labeled $40^\circ$ is $\angle KRA$? No, it looks like $\angle QRA$ or $\angle PRA$? Let's look closer. The angle labeled $40^\circ$ is adjacent to side $R$. The line segment $RA$ connects the vertex $R$ to the incenter $A$. The incenter is the intersection of angle bisectors. Therefore, $RA$ bisects $\angle PRQ$. The angle shown as $40^\circ$ is $\angle KRA$? No, $K$ is on $QR$. The angle is $\angle ARQ = 40^\circ$. Since $RA$ is a bisector, $\angle ARP$ is also $40^\circ$. The question asks for $m\angle ARU$. Point $U$ is on $PR$. So $\angle ARU$ is the same as $\angle ARP$.
* Calculation: Since $RA$ bisects $\angle R$, and half of the angle is $40^\circ$, the other half ($\angle ARU$) is also $40^\circ$.
* Answer: $40^\circ$

8. Find $AU$
* Reasoning: $A$ is the incenter. The incenter is equidistant from all three sides of the triangle. The distance from $A$ to side $PQ$ is given as $AT = 20$. The distance from $A$ to side $PR$ is $AU$. Therefore, $AU = AT$.
* Calculation: $AU = 20$.
* Answer: $20$

9. Find $m\angle UGM$
* Reasoning: $U$ is the incenter of $\triangle GHY$. This means ray $GU$ bisects $\angle HGY$. We are given that $\angle BGY$ (where $B$ is on $GY$) is $28^\circ$. Wait, the angle labeled $28^\circ$ is $\angle UGB$? Or $\angle UGY$? The diagram shows the angle between side $GY$ and the bisector $GU$ is $28^\circ$. Since $GU$ is a bisector, the other half of the angle, $\angle UGH$ (which contains point $M$ on side $GH$), is also $28^\circ$. So, $\angle UGM = 28^\circ$.
* Answer: $28^\circ$

10. Find $m\angle PHU$
* Reasoning: $U$ is the incenter. Ray $HU$ bisects $\angle GHY$. We are given $\angle MHU$? No, we are given $\angle MHU$? Let's look at the top vertex $H$. The angle between side $HY$ and the bisector $HU$ is not directly given. However, we see an angle labeled $21^\circ$ near vertex $H$. It looks like $\angle MHU = 21^\circ$? Or is it $\angle UHY$? The label $21^\circ$ is between side $HM$ (part of $HG$) and the segment $HU$. So $\angle MHU = 21^\circ$. Since $HU$ is an angle bisector, $\angle UHY$ is also $21^\circ$. The question asks for $m\angle PHU$. Point $P$ is on side $HY$. So $\angle PHU$ is the same angle as $\angle UHY$.
* Calculation: Since $HU$ bisects $\angle H$, and $\angle MHU = 21^\circ$, then $\angle PHU = 21^\circ$.
* Answer: $21^\circ$

11. Find $HU$
* Reasoning: We need to find the length of the hypotenuse $HU$ in the right-angled triangle $\triangle HMU$.
* We know $\angle MHU = 21^\circ$ (from problem 10).
* We know the side adjacent to this angle, $HM$? No, we don't have $HM$.
* Let's look at the other information. We have lengths $12$ and $5$.
* The length $12$ is labeled on side $HY$.
* The length $5$ is labeled on segment $UP$. $UP$ is the perpendicular distance from the incenter $U$ to side $HY$.
* Since $U$ is the incenter, it is equidistant from all sides. So, $UM = UP = UB = 5$.
* Now look at $\triangle HMU$. It is a right triangle with right angle at $M$.
* Leg $UM = 5$.
* Angle $\angle MHU = 21^\circ$.
* We want to find the hypotenuse $HU$.
* Use sine ratio: $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$.
* $\sin(21^\circ) = \frac{UM}{HU} = \frac{5}{HU}$.
* $HU = \frac{5}{\sin(21^\circ)}$.
* $\sin(21^\circ) \approx 0.3584$.
* $HU \approx \frac{5}{0.3584} \approx 13.95$.

*Alternative Interpretation:* Is there simpler geometry?
Maybe the "12" is relevant? If $HY = 12$, and $UP=5$, then in $\triangle HPU$ (right angled at $P$):
Hypotenuse $HU = \sqrt{HP^2 + UP^2}$.
We don't know $HP$.

Let's re-read the diagram carefully.
The "12" is next to side $HY$.
The "5" is next to segment $UP$.
$U$ is incenter. $UP \perp HY$, $UM \perp GH$, $UB \perp GY$.
$UP = UM = UB = 5$.

In $\triangle HPU$:
Angle $\angle PHU = 21^\circ$ (since $HU$ bisects $\angle H$ and $\angle MHU=21^\circ$).
Side opposite to $21^\circ$ is $UP = 5$.
We need hypotenuse $HU$.
$\sin(21^\circ) = \frac{5}{HU}$.
$HU = \frac{5}{\sin(21^\circ)}$.

Let's calculate:
$5 / 0.35836... = 13.952...$

Rounding to nearest tenth: $14.0$.
Rounding to nearest whole number: $14$.

Let's check if there is another way. What if the triangle is special? No obvious special angles.
What if I use cosine?
In $\triangle HPU$, adjacent side is $HP$.
$\tan(21^\circ) = \frac{5}{HP} \Rightarrow HP = \frac{5}{\tan(21^\circ)} \approx \frac{5}{0.3838} \approx 13.02$.
Then $HU = \sqrt{13.02^2 + 5^2} = \sqrt{169.5 + 25} = \sqrt{194.5} \approx 13.95$.

The result is consistent.

* Answer: $\approx 14$ (or $13.95$)

Final Answer:
1. $43^\circ$
2. $47^\circ$
3. $19$
4. $33^\circ$
5. $7$
6. $11^\circ$
7. $40^\circ$
8. $20$
9. $28^\circ$
10. $21^\circ$
11. $14$