1. Segment line AD bisects Angle BAC. If BD=DA-AC, what is Angle C in degrees?
2. In the diagram below, I is the incenter of Triangle ABC. We know that AB=14, BC=11, and CA=15. What is CE?
3. In the diagram below, I is the incenter of Triangle ABC. The segment line DE goes through I and is parallel to line AC. If AE=4 and CD-3, what is DE?
4. As shown in the diagram Angle BAC=90 degrees. Let X be the foot of the altitude from A to line BC, line AY be the bisector of Angle BAC, and line AZ be a median of Triangle ABC. If Angle XAY=13 degrees and X is on line BY, then what is the measure of angle ZAC in degrees?
5. I is the incenter of Triangle ABC as shown below. We have CE=15,AE=9 , and AB=12. Find BF.
6. I is the incenter of Triangle XYZ. A circle centered at I intersects the three sides of Triangle XYZ at the six points A,B C,D E,and F as shown below. We know that YZ=4, ZX=5, and XY=6=AB+CD+EF. Find XA.
1. Segment line AD bisects Angle BAC. If BD=DA-AC, what is Angle C in degrees?
I think you must mean that BD = DA = AC
[ DA - AC couldn't be as long as BD ]
Since BD = AD, then angles BAD and ABD are equal
And angle CAD = angle BAD
And since DA = AC, then angle ACD = angle ADC
Let the measures of angles ABD, BAD, CAD = x
And let the measures of angles ACD and ADC = y
So we have that
x + x + x + y = 180 ⇒ 3x + y = 180 ⇒ y = 180 -3x (1)
x + y + y = 180 ⇒ x + 2y = 180 (2)
Sub (1) into (2)
x + 2(180 - 3x) = 180
x + 360 - 6x = 180
-5x = - 180
x = 36°
And angle ACD = angle C = 180 - 3(36) = 180 - 108 = 72°
2. In the diagram below, I is the incenter of Triangle ABC. We know that AB=14, BC=11, and CA=15. What is CE?
Draw CI, AI and BI
Because of AAS......
Triangles EIC and DIC are congruent and
Triangles EIA and FIA are congruent and
Triangles DIB and FIB are congruent
So
EC = DC = x and
EA = FA = y and
BD = BF = z
So we have this system of equations
x + y = 15 ⇒ y = 15 - x (1)
y + z = 14 (2)
x + z = 11 ⇒ z = 11 - x (3)
Sub (1) and (3) into (2)
15 - x + 11 - x = 14 simplify
26 - 2x = 14
-2x = - 12
x = 6 = CE
5. I is the incenter of Triangle ABC as shown below. We have CE=15,AE=9 , and AB=12. Find BF.
Because BE is an angle bisector....we have the following relationship
BC / AB = CE /AE
BC / 12 = 15 / 9
BC /12 = 5/3
BC = 12 * 5 / 3 = 20
And because FC is an angle bisector, we have this relationship
BC /AC = BF /AF
20 / 24 = BF / AF
5 / 6 = BF / AF
So.....AC is divided into 11 equal parts......and BF is 5 of these
So BF is (5/11) of AC = (5/11) * 12 = 60 / 11
3. In the diagram below, I is the incenter of Triangle ABC. The segment line DE goes through I and is parallel to line AC. If AE=4 and CD-3, what is DE?
Because IC is a transversal cutting parallel lines, then angles ICA and DIC are equal alternate angles......but angles ICA and ICD are equal because IC is an angle bisector
Therefore, angles DIC and ICD are equal
But.......in triangle DIC, the sides opposite these angles are also equal
So ....CD = ID = 3
Similarly, we can show that AE = IE = 4
So DE = DI + EI = 3 + 4 = 7
4.
As shown in the diagram Angle BAC=90 degrees.
Let X be the foot of the altitude from A to line BC,
line AY be the bisector of Angle BAC,
and line AZ be a median of Triangle ABC.
If Angle XAY=13 degrees and X is on line BY,
then what is the measure of angle ZAC in degrees?
\(\text{Let $BZ=CZ=AZ=r$} \\ \text{Let $ZAC=x$} \\ \text{Let $ZAC=ACZ=x$}\)
\(\begin{array}{|rcll|} \hline ABC &=& 90^{\circ}-ACZ\\ &=& 90^{\circ}-x \\\\ BAX &=& 90^{\circ}-(90^{\circ}-x) \\ &=& x \\ \hline \end{array}\)
\(\text{Let $BAY=YAC$} \)
\(\begin{array}{|rcll|} \hline BAY &=& x+13^{\circ} \\ BAY=YAC &=& x+13^{\circ} \\ \mathbf{YAZ} & \mathbf{=} & \mathbf{13^{\circ} } \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline AYX &=& 90^{\circ} - 13^{\circ} \\ &=& 77^{\circ} \\\\ CYA &=& 180^{\circ} - AYX \\ &=& 180^{\circ} - 77^{\circ} \\ &=& 103^{\circ} \\ \hline \end{array}\)
\(\text{Let $AZY=2x$} \)
\(\begin{array}{|rcll|} \hline AZY+CYA+YAZ &=& 180^{\circ} \\ 2x+103^{\circ} +13^{\circ} &=& 180^{\circ} \\ 2x &=& 180^{\circ} -103^{\circ} -13^{\circ} \\ 2x &=& 64^{\circ} \\ \mathbf{x} &\mathbf{=}& \mathbf{32^{\circ}} \\ \hline \end{array}\)
The measure of angle ZAC in degrees is \(\mathbf{32^{\circ}}\)
6. I is the incenter of Triangle XYZ. A circle centered at I intersects the three sides of Triangle XYZ at the six points A,B C,D E,and F as shown below. We know that YZ=4, ZX=5, and XY=6=AB+CD+EF. Find XA.
Draw IX, IY, IZ
Draw IM perpendicular to YZ
Draw IN perpendicular to XZ
Draw IO perpendicular to XY
By AAS
Triangle IMZ congruent to Triangle INZ
Triangle INX concruent to Triangle IOX
Triangle IMY congruent to Triangle IOY
So.....
IM = IN = IO
But....chords equidistant from the center of a circle are themselves equal
So AB = CD = EF
So
AB + CD + EF = 6 and by substitution, we have that
AB + AB + AB = 6
3AB = 6
AB = 2 = CD = EF
So
AB/2 = CD/2 = EF/2
OY = MY
AB/2 + BY = CD/2 + CY
OX = NX
AB/2 + AX = EF/2 + FX
NX = MX
EF/2 + EZ = CD/2 + DZ
So......this implies that
CY = BY = a
AX = FX = b
EZ = DZ = c
So......we have this system
a + CD + c = 4 ⇒ a + c = 2 ⇒ a = 2 - c (1)
a + AB + b = 6 ⇒ a + b = 4 (2)
b + EF + c = 5 ⇒ c + b = 3 ⇒ b = 3 - c (3)
Sub (1) and (3) into (2)
2 - c + 3 - c = 4
5 - 2c = 4
-2c = -1
c = 1/2
b = XA = 3 - c = 3 - 1/2 = 5/2