Question

multiple choice (2 pts. each)
#1.) in △ abc, m∠a = (2x)°, m∠b = (3x − 25)°, and m∠c = (x + 55)°. what is the type of triangle is △ abc?
2x + 3x - 25 + x + 55 = 180
(a) equilateral
(b) isosceles 6x + 30 = 180
(c) right - 30 - 30
(d) scalene 6x = 150
x = 25

use the partial proof below to determine what goes in the blanks for questions 2-5.
given: ( overline{ae} cong overline{bc} ), d is the midpoint of ( overline{ce} ), ( overline{ce} perp overline{cb} ), and ( overline{ec} perp overline{ea} )
prove: ( \triangle aed cong \triangle bcd )

statement	reason
#2.	2) midpoint definition
3) ( mangle e = 90^circ ), ( mangle c = 90^circ )	3) #3.
4) ( angle e cong angle c )	4) #4.
5) ( \triangle aed cong \triangle bcd )	5) #5.

*#2.)
(a) ( ae cong bc )
(b) ( de cong dc )
(c) ( ad cong bd )
(d) ( de cong ae )

**#3.)
(a) definition of right angles
(b) perpendicular line definition
(c) all right angles are congruent
(d) midpoint definition

**#4.)
(a) definition of right angles
(b) perpendicular line definition
(c) all right angles are congruent
(d) midpoint definition

#5.)
(a) sas
(b) ssa
(c) asa
(d) aas

Explanation:

Question 1:

Step1: Sum of triangle angles

The sum of the interior angles of a triangle is \(180^\circ\). So, \(m\angle A + m\angle B + m\angle C = 180^\circ\). Substitute the given angles: \((2x)^\circ+(3x - 25)^\circ+(x + 55)^\circ = 180^\circ\).

Step2: Solve for \(x\)

Simplify the left - hand side: \(2x+3x - 25+x + 55=180\), which combines to \(6x + 30 = 180\). Subtract 30 from both sides: \(6x=180 - 30=150\). Then divide by 6: \(x = \frac{150}{6}=25\).

Step3: Find each angle

• \(m\angle A=2x = 2\times25 = 50^\circ\)
• \(m\angle B=3x - 25=3\times25-25 = 50^\circ\)
• \(m\angle C=x + 55=25 + 55 = 80^\circ\)

Since two angles (\(\angle A\) and \(\angle B\)) are equal, the triangle has two equal sides (by the isosceles triangle theorem: if two angles of a triangle are equal, then the sides opposite those angles are equal). So the triangle is isosceles.

The midpoint definition states that if \(D\) is the midpoint of \(\overline{CE}\), then \(DE = DC\). So the correct statement for #2 is \(DE\cong DC\), which corresponds to option B.

We know that \(\overline{CE}\perp\overline{EB}\) and \(\overline{EC}\perp\overline{EA}\). By the definition of perpendicular lines, if two lines are perpendicular, they form right angles (angles with measure \(90^\circ\)). So the reason for \(m\angle E = 90^\circ\) and \(m\angle C=90^\circ\) is the Perpendicular Line Definition, which is option B.

Answer:

B. Isosceles

Question 2: