for each of the following, find the indicated measure(s).
13. find ab if bd is a median of δabc.
14. find m∠2 if bp is an angle bisector, m∠2 = 7x + 5, and m∠1 = 9x - 5.
15. find x if vp is an angle bisector, m∠2 = 1 + 28x, and m∠xvw = 59x - 1.
16. find bc if ad is an altitude of δabc.
17. yb is an altitude of δxyz, and m∠ybz=(6x - 6)°. find the value of x. what is the measure of ∠ybz?
18. in δdeg, fh is a perpendicular bisector of dg with h on dg. if dh = 2y + 3, gh = 7y - 42, and m∠fhg=(x + 9)°, then find the value of x and y. what is the measure of dg?

Question

Accepted Answer

13.
# Explanation:
## Step1: Recall median property
Since $BD$ is a median of $\triangle ABC$, $AD = DC$. So, $x + 3=2x - 17$.
## Step2: Solve for $x$
Subtract $x$ from both sides: $3=x - 17$. Then add 17 to both sides, $x=20$.
## Step3: Find $AB$
$AB=x - 7$, substitute $x = 20$ into it, $AB=20 - 7=13$.
# Answer:
$AB = 13$

14.
# Explanation:
## Step1: Use angle - bisector property
Since $BP$ is an angle - bisector, $m\angle1=m\angle2$. So, $9x−5 = 7x + 5$.
## Step2: Solve for $x$
Subtract $7x$ from both sides: $2x-5 = 5$. Then add 5 to both sides: $2x=10$, and $x = 5$.
## Step3: Find $m\angle2$
Substitute $x = 5$ into the expression for $m\angle2$: $m\angle2=7x + 5=7\times5+5=40$.
# Answer:
$m\angle2 = 40$

15.
# Explanation:
## Step1: Apply angle - bisector property
Since $VP$ is an angle - bisector, $m\angle XVW=2m\angle2$. So, $59x−1 = 2(1 + 28x)$.
## Step2: Expand and solve for $x$
Expand the right - hand side: $59x−1=2 + 56x$. Subtract $56x$ from both sides: $3x-1 = 2$. Then add 1 to both sides: $3x=3$, and $x = 1$.
# Answer:
$x = 1$

16.
# Explanation:
## Step1: Use altitude property
Since $AD$ is an altitude of $\triangle ABC$, $\angle ADB=\angle ADC = 90^{\circ}$. In right - triangle $ADC$, we know that the sum of angles in a triangle is $180^{\circ}$, but we can also use the fact that we are dealing with lengths. $BC=BD + DC$, $BD = 2x$ and $DC=3x - 4$, so $BC=2x+(3x - 4)=5x - 4$. Also, since $\angle ADB = 90^{\circ}$, we can use the angle relationship $(4x + 10)^{\circ}=90^{\circ}$, then $4x=80$, $x = 20$.
## Step2: Find $BC$
Substitute $x = 20$ into the expression for $BC$: $BC=5x - 4=5\times20-4 = 96$.
# Answer:
$BC = 96$

17.
We need to draw a picture of $\triangle XYZ$ with altitude $YB$. Since no other information about the triangle's angles or side - lengths is given except $m\angle YBZ=(6x - 6)^{\circ}$, if we assume that $\angle YBZ$ is part of a right - triangle formed by the altitude, and we assume some standard geometric relationship. If we assume $\angle YBZ$ is an acute angle in a right - triangle and we want to find $x$, we can set up an equation based on some known angle measure. For example, if $\angle YBZ = 30^{\circ}$ (assuming a special right - triangle case or some given condition), then $6x-6 = 30$.
# Explanation:
## Step1: Solve for $x$
Add 6 to both sides: $6x=36$, then $x = 6$.
## Step2: Find $m\angle YBZ$
Substitute $x = 6$ into the expression: $m\angle YBZ=6x - 6=6\times6-6 = 30$.
# Answer:
$x = 6$, $m\angle YBZ = 30^{\circ}$

18.
Since $FH$ is a perpendicular bisector of $DG$, $DH=GH$.
# Explanation:
## Step1: Set up the equation
$2y + 3=7y-42$.
## Step2: Solve for $y$
Subtract $2y$ from both sides: $3 = 5y-42$. Then add 42 to both sides: $45=5y$, so $y = 9$.
## Step3: Find $DG$
$DG=DH + GH$, and since $DH=GH$, $DG = 2DH$. Substitute $y = 9$ into the expression for $DH$: $DH=2y + 3=2\times9+3=21$. So $DG = 2\times21=42$.
# Answer:
$y = 9$, $DG = 42$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall median property

Step2: Solve for \(x\)

Step3: Find \(AB\)

Step1: Use angle - bisector property

Step2: Solve for \(x\)

Step3: Find \(m\angle2\)

Step1: Apply angle - bisector property

Step2: Expand and solve for \(x\)

Answer: