if the triangle abc shown, dc=20, ab=21, de=12, and ab is parallel to de. what is the value of ac
ane. Segment line Advertizing 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-three, what is DE?
4. Every bit shown in the diagram Angle BAC=xc degrees. Let X be the foot of the altitude from A to line BC, line AY exist the bisector of Bending 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 mensurate of bending ZAC in degrees?
v. I is the incenter of Triangle ABC as shown below. Nosotros accept CE=xv,AE=9 , and AB=12. Observe 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.
+121433 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 = Advertisement, then angles BAD and ABD are equal
And angle CAD = bending BAD
And since DA = AC, then angle ACD = bending ADC
Let the measures of angles ABD, BAD, CAD = ten
And permit the measures of angles ACD and ADC = y
And so we have that
ten + ten + x + y = 180 ⇒ 3x + y = 180 ⇒ y = 180 -3x (1)
x + y + y = 180 ⇒ x + 2y = 180 (2)
Sub (one) into (2)
x + 2(180 - 3x) = 180
ten + 360 - 6x = 180
-5x = - 180
x = 36°
And angle ACD = angle C = 180 - 3(36) = 180 - 108 = 72°
+121433 2. In the diagram below, I is the incenter of Triangle ABC. Nosotros know that AB=xiv, BC=11, and CA=fifteen. What is CE?
Describe CI, AI and BI
Considering of AAS......
Triangles EIC and DIC are congruent and
Triangles Environmental impact assessment and FIA are congruent and
Triangles DIB and FIB are congruent
Then
EC = DC = ten and
EA = FA = y and
BD = BF = z
So we have this system of equations
10 + y = fifteen ⇒ y = fifteen - x (i)
y + z = 14 (2)
x + z = xi ⇒ z = 11 - x (3)
Sub (1) and (3) into (2)
fifteen - ten + 11 - ten = fourteen simplify
26 - 2x = 14
-2x = - 12
x = 6 = CE
+121433 5. I is the incenter of Triangle ABC as shown below. We accept CE=xv,AE=nine , and AB=12. Find BF.
Because Be is an angle bisector....we have the post-obit relationship
BC / AB = CE /AE
BC / 12 = 15 / ix
BC /12 = 5/iii
BC = 12 * v / 3 = 20
And because FC is an angle bisector, we have this relationship
BC /AC = BF /AF
xx / 24 = BF / AF
5 / six = BF / AF
So.....Air conditioning is divided into 11 equal parts......and BF is v of these
And so BF is (5/eleven) of AC = (5/11) * 12 = 60 / eleven
+121433 iii. In the diagram below, I is the incenter of Triangle ABC. The segment line DE goes through I and is parallel to line Air conditioning. If AE=4 and CD-3, what is DE?
Because IC is a transversal cutting parallel lines, then angles ICA and DIC are equal alternating 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 reverse these angles are also equal
So ....CD = ID = 3
Similarly, we can show that AE = IE = 4
And then DE = DI + EI = 3 + 4 = 7
+26236 4.
Every bit shown in the diagram Angle BAC=90 degrees.
Permit X exist the pes of the distance 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=thirteen degrees and Ten is on line BY,
then what is the measure of bending ZAC in degrees?
\(\text{Allow $BZ=CZ=AZ=r$} \\ \text{Let $ZAC=x$} \\ \text{Let $ZAC=ACZ=x$}\)
\(\begin{array}{|rcll|} \hline ABC &=& 90^{\circ}-ACZ\\ &=& 90^{\circ}-10 \\\\ BAX &=& ninety^{\circ}-(90^{\circ}-x) \\ &=& 10 \\ \hline \cease{array}\)
\(\text{Let $BAY=YAC$} \)
\(\brainstorm{array}{|rcll|} \hline BAY &=& x+13^{\circ} \\ BAY=YAC &=& x+thirteen^{\circ} \\ \mathbf{YAZ} & \mathbf{=} & \mathbf{13^{\circ} } \\ \hline \cease{array} \)
\(\brainstorm{array}{|rcll|} \hline AYX &=& 90^{\circ} - xiii^{\circ} \\ &=& 77^{\circ} \\\\ CYA &=& 180^{\circ} - AYX \\ &=& 180^{\circ} - 77^{\circ} \\ &=& 103^{\circ} \\ \hline \cease{array}\)
\(\text{Let $AZY=2x$} \)
\(\begin{assortment}{|rcll|} \hline AZY+CYA+YAZ &=& 180^{\circ} \\ 2x+103^{\circ} +xiii^{\circ} &=& 180^{\circ} \\ 2x &=& 180^{\circ} -103^{\circ} -13^{\circ} \\ 2x &=& 64^{\circ} \\ \mathbf{x} &\mathbf{=}& \mathbf{32^{\circ}} \\ \hline \end{array}\)
The measure of angle ZAC in degrees is \(\mathbf{32^{\circ}}\)
+121433 Very well washed, Heureka.....!!!
+121433 6. I is the incenter of Triangle XYZ. A circumvolve centered at I intersects the 3 sides of Triangle XYZ at the six points A,B C,D E,and F every bit shown below. We know that YZ=4, ZX=5, and XY=half dozen=AB+CD+EF. Find XA.
Draw IX, IY, IZ
Draw IM perpendicular to YZ
Depict 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
Merely....chords equidistant from the heart of a circle are themselves equal
So AB = CD = EF
And then
AB + CD + EF = 6 and by exchange, we have that
AB + AB + AB = 6
3AB = 6
AB = ii = CD = EF
So
AB/2 = CD/2 = EF/two
OY = MY
AB/two + Past = CD/two + CY
OX = NX
AB/two + AX = EF/ii + FX
NX = MX
EF/2 + EZ = CD/ii + DZ
So......this implies that
CY = Past = a
AX = FX = b
EZ = DZ = c
And then......nosotros 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 = five ⇒ c + b = 3 ⇒ b = three - c (iii)
Sub (1) and (3) into (2)
2 - c + 3 - c = four
5 - 2c = 4
-2c = -one
c = ane/two
b = XA = 3 - c = 3 - one/2 = 5/2
delight requite me a like. I am poor. lol
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