Activity 2 Direction: Complete the proof. Given: V is the midpoint of |E, angle| cong angle E Prove: V is the midpoint of GR Proof: Statements 1. V is the midpoint of IE, 21 2. 3. 4. AGVI ARVE 5. 6. V is the midpoint of GR Q3 Week No.9 Competency E R Reasons Definition of Midpoint E 1. 2. 3. Vertical Angle Theorem 5. CPCTC 6. 5 Code: M8GE-Illh-1, M8GE-Illi-j-1 Activity 3 Direction: Fill in the missing statements and reasons. Given: GA bisects angle G; overline GA perp L overline D Prove: overline GL cong overline GD Proof: Statements 1. GA bisects 4G 2 . 3 . 4. ZGAL and angle GAD are right angles 5 . 6. overline GA cong overline GA; 7. Delta GAL cong Delta GAD; 8. overline GL cong overline GD L D Reasons Bisector 1. 2. Definition of Angle 3. Given 4 . 5. All right angles are congruent. 6. 7 . 8.​

Step-by-step explanation:

Given

See attachment for right question format

Required

Complete the blanks

1. V is the of |E, ---- This statement was stated in the question.

So, the reason is Given

2. Midpoint means halfway. So, we can fill the statement column with:

--- because V is the midpoint of IV and VE

or --- because V is the midpoint of GV and VR

3. Vertical angle are congruent. So, the statement is:

which means that angles at V are congruent

4. We have:

because

--- congruent angles (A)

---- congruent sides (S)

 --- congruent angles (A)

The above implies that: because of ASA congruence

5. CPCTC is true here because GV and VR are corresponding sides of triangles GVI and RVE. ----

Hence, both sides are congruent ----

6. The given statement has been prived

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.