23) an altitude is drawn from vertex ∠y to the opposite side xz at point w. what is the relationship between yw and xz.
24) dh and eg bisect each other, which of the triangle congruence methods could be used to prove △def ≅ △hgf?
a. sss
b. sas
c. asa
d. aas
25) given the two congruent triangles below, find the values of x and y.
equation ______________ equation: ______________
x = ______ y = ______
26) if e bisects of ab and ed ≅ ec, what additional information is needed to prove that △aed ≅ △bec by the hl congruence theorem?

Question

Accepted Answer

### Question 23
# Explanation:
## Step1: Recall the definition of an altitude
An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. So, $ \overline{YW} $ is an altitude from $ Y $ to $ \overline{XZ} $, which means $ \angle YWX $ and $ \angle YWZ $ are right angles ($ 90^\circ $).
## Step2: Determine the relationship
Since $ \overline{YW} $ is perpendicular to $ \overline{XZ} $, the relationship between $ \overline{YW} $ and $ \overline{XZ} $ is that they are perpendicular. In other words, $ YW \perp XZ $.
# Answer: $ YW $ is perpendicular to $ XZ $ (or $ YW \perp XZ $)

### Question 24
# Brief Explanations:
1. Given that $ \overline{DH} $ and $ \overline{EG} $ bisect each other at $ F $, so $ DF = FH $ and $ EF = FG $ (by the definition of bisecting a segment).
2. Also, $ \angle DFE $ and $ \angle HFG $ are vertical angles, so they are congruent ($ \angle DFE \cong \angle HFG $).
3. Now, in $ \triangle DEF $ and $ \triangle HGF $, we have $ DF = FH $, $ \angle DFE \cong \angle HFG $, and $ EF = FG $. This satisfies the SAS (Side - Angle - Side) congruence criterion (two sides and the included angle of one triangle are congruent to two sides and the included angle of the other triangle).
4. So, the triangle congruence method that can be used to prove $ \triangle DEF \cong \triangle HGF $ is SAS.
# Answer: B. SAS

### Question 25
# Explanation:
## Step1: Find the equation for $ x $
Since the triangles are congruent, their corresponding sides are equal. The side with length $ 8x - 3 $ corresponds to the side with length $ 22 $. So we set up the equation:
$ 8x - 3=22 $
## Step2: Solve for $ x $
Add 3 to both sides of the equation:
$ 8x=22 + 3=25 $
Divide both sides by 8:
$ x=\frac{25}{8}=3.125 $
## Step3: Find the equation for $ y $
The side with length $ 6y + 1 $ corresponds to the side with length $ 29 $. So we set up the equation:
$ 6y+1 = 29 $
## Step4: Solve for $ y $
Subtract 1 from both sides:
$ 6y=29 - 1 = 28 $
Divide both sides by 6:
$ y=\frac{28}{6}=\frac{14}{3}\approx4.666\cdots $ (or as a fraction $ \frac{14}{3} $)
# Answer:
Equation for $ x $: $ 8x - 3 = 22 $
Equation for $ y $: $ 6y + 1 = 29 $
$ x=\frac{25}{8}=3.125 $
$ y=\frac{14}{3}\approx4.67 $ (or $ \frac{14}{3} $)

### Question 26
# Brief Explanations:
1. The HL (Hypotenuse - Leg) Congruence Theorem states that if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent.
2. We know that $ E $ bisects $ \overline{AB} $, so $ AE = EB $. Also, $ \overline{ED} \cong \overline{EC} $.
3. For $ \triangle AED $ and $ \triangle BEC $ to be right triangles (required for HL), we need to know that $ \angle A $ and $ \angle B $ are right angles (since $ ABCD $ looks like a rectangle in the diagram, so $ \angle A $ and $ \angle B $ should be $ 90^\circ $). Alternatively, we need to know that $ AD \perp AB $ and $ BC \perp AB $ (so that $ \triangle AED $ and $ \triangle BEC $ are right triangles with right angles at $ A $ and $ B $ respectively).
4. So the additional information needed is that $ \angle A $ and $ \angle B $ are right angles (or $ AD \perp AB $ and $ BC \perp AB $, or $ ABCD $ is a rectangle, etc.). A more precise answer is that $ \angle A $ and $ \angle B $ are right angles (i.e., $ \angle A = 90^\circ $ and $ \angle B = 90^\circ $) or that $ AD \perp AB $ and $ BC \perp AB $.
# Answer: The additional information needed is that $ \angle A $ and $ \angle B $ are right angles (or $ AD \perp AB $ and $ BC \perp AB $, or $ ABCD $ is a rectangle, etc. - a common answer is that $ \angle A = 90^\circ $ and $ \angle B = 90^\circ $)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 23

Step1: Recall the definition of an altitude

Step2: Determine the relationship

Step1: Find the equation for \( x \)

Step2: Solve for \( x \)

Step3: Find the equation for \( y \)

Step4: Solve for \( y \)

Answer:

Question 24