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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. Nov 26, 2018

#1
+2

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 ]

And since DA = AC, then angle  ACD = angle ADC

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°   Nov 26, 2018
#2
+2

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   Nov 26, 2018
#3
+2

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   Nov 26, 2018
#4
+2

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   Nov 26, 2018
#5
+10

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 $$\mathbf{32^{\circ}}$$ Nov 26, 2018
#6
+1

Very well done, Heureka.....!!!   CPhill  Nov 26, 2018
#7
+1

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   Nov 26, 2018
#8
+2

please give me a like. I am poor. lol

Nov 26, 2018
edited by Guest  Nov 26, 2018