Question

complete the statement for the diagram at the right.

1. ? is bisected by ?.
2. the measure of ∠abc is ? the measure of ∠abd.

ge bisects ∠fgh. find the angle measure.

1. find m∠egh.
2. find m∠fge.
3. find m∠egf.

ba bisects ∠dbc. find m∠cba and m∠dbc.
6.
7.
8.
jk bisects ∠gjh. find the value of x.
9.
10.
11.
use the diagram at the right. decide whether the statement is true or false.

1. if bd bisects ∠abc, then ∠abd≅∠abc.
2. if bd bisects ∠abc, then ∠dbc≅∠abd.
3. if bd bisects ∠abc and m∠abd = 55°, then m∠dbc = 55°.
4. if bd bisects ∠abc and m∠abc = 112°, then m∠abd = 61°.
5. in the pup tent shown at the right, the two sides meet at the top to form a 72° angle. if the tent pole bisects the angle, what angle does the tent pole make with each of the sides?

Explanation:

Step1: Recall angle - bisector definition

An angle - bisector divides an angle into two congruent angles.

Step2: Solve problem 3

Since $\overrightarrow{GE}$ bisects $\angle FGH$ and $\angle FGE = 25^{\circ}$, then $\angle EGH=\angle FGE = 25^{\circ}$.

Step3: Solve problem 4

Since $\overrightarrow{GE}$ bisects $\angle FGH$ and $\angle FGH=140^{\circ}$, then $\angle FGE=\frac{1}{2}\angle FGH=\frac{1}{2}\times140^{\circ}=70^{\circ}$.

Step4: Solve problem 5

If $\angle FGH = 90^{\circ}$ and $\overrightarrow{GE}$ bisects $\angle FGH$, then $\angle EGF=\frac{1}{2}\angle FGH = 45^{\circ}$.

Step5: Solve problem 6

Since $\overrightarrow{BA}$ bisects $\angle DBC$ and $\angle DBA = 35^{\circ}$, then $\angle CBA=\angle DBA = 35^{\circ}$ and $\angle DBC = 2\angle DBA=70^{\circ}$.

Step6: Solve problem 7

Since $\overrightarrow{BA}$ bisects $\angle DBC$ and $\angle DBA = 75^{\circ}$, then $\angle CBA=\angle DBA = 75^{\circ}$ and $\angle DBC = 2\angle DBA = 150^{\circ}$.

Step7: Solve problem 8

Since $\overrightarrow{BA}$ bisects $\angle DBC$ and $\angle DBA = 45^{\circ}$, then $\angle CBA=\angle DBA = 45^{\circ}$ and $\angle DBC = 2\angle DBA=90^{\circ}$.

Step8: Solve problem 9

Since $\overrightarrow{JK}$ bisects $\angle GJH$, then $\angle GJK=\angle KJH$. So $x + 10=40$, and $x=30$.

Step9: Solve problem 10

Since $\overrightarrow{JK}$ bisects $\angle GJH$, then $4x=\frac{1}{2}\times80$, so $4x = 40$ and $x = 10$.

Step10: Solve problem 11

Since $\overrightarrow{JK}$ bisects $\angle GJH$, then $18+2x=6x$. Subtracting $2x$ from both sides gives $18 = 4x$, so $x = 4.5$.

Step11: Solve problem 12

If $\overrightarrow{BD}$ bisects $\angle ABC$, then $\angle ABD=\frac{1}{2}\angle ABC$, so the statement “$\angle ABD\cong\angle ABC$” is False.

Step12: Solve problem 13

If $\overrightarrow{BD}$ bisects $\angle ABC$, then $\angle DBC=\angle ABD$ by the definition of an angle - bisector, so the statement is True.

Step13: Solve problem 14

If $\overrightarrow{BD}$ bisects $\angle ABC$ and $\angle ABD = 55^{\circ}$, then $\angle DBC=\angle ABD = 55^{\circ}$ by the definition of an angle - bisector, so the statement is True.

Step14: Solve problem 15

If $\overrightarrow{BD}$ bisects $\angle ABC$ and $\angle ABC = 112^{\circ}$, then $\angle ABD=\frac{1}{2}\angle ABC=56^{\circ}
eq61^{\circ}$, so the statement is False.

Step15: Solve problem 16

If the tent pole bisects a $72^{\circ}$ angle, then the angle it makes with each side is $\frac{1}{2}\times72^{\circ}=36^{\circ}$.

Answer:

1. $25^{\circ}$
2. $70^{\circ}$
3. $45^{\circ}$
4. $m\angle CBA = 35^{\circ},m\angle DBC = 70^{\circ}$
5. $m\angle CBA = 75^{\circ},m\angle DBC = 150^{\circ}$
6. $m\angle CBA = 45^{\circ},m\angle DBC = 90^{\circ}$
7. $x = 30$
8. $x = 10$
9. $x = 4.5$
10. False
11. True
12. True
13. False
14. $36^{\circ}$