5. Given: D is the midpoint of BC of triangle ABC Prove: EF//BC Statements Reasons 1.D is the midpoint of side BC of triangle 1.Given ABC and the bisectors of angles ADB and ADC meet AB and AC at E and F respectively 3.Triangle ABC = triangle AEF 3.lf two angles of one triangle are equal respectively to two angles of another, then the triangle are similar. (a.a.) 4.AE + EB = AB & AF+FC = AC 4.Segment Addition Postulate %3D 5.Triangle BDE = triangle ADE & triangle CDF = triangle ADF 5.Definition of angle bisector %3D 6.Corresponding sides of similar triangles are proportional (C.S.S.T.P.) 6.AE/EB = AF/FC %3D 7.Angle ABD = angle AEF & angle BCA = 7.Corresponding Angles Postulate angle EFA %3D 8.DE bisects AB and DF bisects AC 8. proportionally 9. If a line divides two sides of a triangle proportionally, then it is parallel to the third side. (Theorem 54) 9.EF // BC ET !! AutoSave Exercises 4_15 proofs 1 and. . Marc Skwarczynski ON Exercise 4.15 #3: A triangle ABC is inscribed in a circle. M is the midpoint of AMČ and BM intersects AC at D. File Home Insert Draw Design Layout References Mailings Review View Help Table Design Layout 1D Exercises 4.15 #3 Given: Inscribed AABC; M midpoint of AMC; BM intersects AC in D AB AD Prove: %3D BC DC AB Prove: BC AD Statements Reasons DC 1. (see above) 1. Given 2. CM = AM 2. Def. of midpoint 3. ZCBM = - CM; 3. An inscribed angle = ½ the intercepted arc. %3D %3D ZABM = AM M. 4. - CM = - AM 4. Division (of step 2) %3D 2. 5. Substitution 5. ZCBM = LABM %3D (step 3 -> step 4) 6. Def. of bisector 6. BM bisects ZABC 7. The bisector of an interior angle of a divides the oppos internally into se which have the sa AB 7. BC AD %3D DC as the other two 257 words fecus

Question

Please help with this two-column proof. Just fill in the statements and reasons in the missing blanks. *There is an example of how it looks like when it is done in the second picture.*

The proof is in the picture. **This is about proving the triangle true, not about writing essay.**

*Maybe, if you think those reasons below suit the statements, you can use it.* (Those below are the reasons you can use to prove the statements true if you think it is suited for it.)

Definition of ~ triangles

Definition of bisector

Definition of perimeter

Definition of median

Definition of midpoint

Multiplication

Division

An inscribed angle = 1/2 the intercepted arc

If 4 quantities are in proportion, then like powers are in proportion.

Subtraction Transformation

Alternation Transformation

Pythagorean Theorem

If 2 angles have their sides perpendicular, right side to right side and left side to left side, the angles are equal.

In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent.

In any proportion the product of the means equals the product of the extremes.

Angles inscribed in the same segment or equal segments are equal.

If a line is drawn parallel to the base of a triangle, it cuts off a triangle similar to the given triangle.

Two isosceles triangles are similar if any angle of one equals the corresponding angle of the other.

5.
Given: D is the midpoint of BC of triangle ABC
Prove: EF//BC
Statements
Reasons
1.D is the midpoint of side BC of triangle 1.Given
ABC and the bisectors of angles ADB
and ADC meet AB and AC at E and F
respectively
3.Triangle ABC = triangle AEF
3.lf two angles of one triangle are equal
respectively to two angles of another,
then the triangle are similar. (a.a.)
4.AE + EB = AB & AF+FC = AC
4.Segment Addition Postulate
%3D
5.Triangle BDE = triangle ADE & triangle
CDF = triangle ADF
5.Definition of angle bisector
%3D
6.Corresponding sides of similar
triangles are proportional (C.S.S.T.P.)
6.AE/EB = AF/FC
%3D
7.Angle ABD = angle AEF & angle BCA = 7.Corresponding Angles Postulate
angle EFA
%3D
8.DE bisects AB and DF bisects AC
8.
proportionally
9. If a line divides two sides of a triangle
proportionally, then it is parallel to the
third side. (Theorem 54)
9.EF // BC
ET
!!
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AutoSave
Exercises 4_15 proofs 1 and. .
Marc Skwarczynski
ON
Exercise 4.15 #3: A triangle ABC
is inscribed in a circle. M is the
midpoint of AMČ and BM
intersects AC at D.
File
Home
Insert
Draw
Design
Layout
References
Mailings
Review
View
Help
Table Design
Layout
1D
Exercises 4.15 #3
Given: Inscribed AABC; M midpoint of AMC; BM
intersects AC in D
AB AD
Prove:
%3D
BC
DC
AB
Prove:
BC
AD
Statements
Reasons
DC
1. (see above)
1. Given
2. CM = AM
2. Def. of midpoint
3. ZCBM = - CM;
3. An inscribed angle = ½
the intercepted arc.
%3D
%3D
ZABM =
AM
M.
4. - CM = - AM
4. Division (of step 2)
%3D
2.
5. Substitution
5. ZCBM = LABM
%3D
(step 3 -> step 4)
6. Def. of bisector
6. BM bisects ZABC
7. The bisector of an
interior angle of a
divides the oppos
internally into se
which have the sa
AB
7.
BC
AD
%3D
DC
as the other two
257 words
fecus
View transcribed image text
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Transcribed Image Text

5. Given: D is the midpoint of BC of triangle ABC Prove: EF//BC Statements Reasons 1.D is the midpoint of side BC of triangle 1.Given ABC and the bisectors of angles ADB and ADC meet AB and AC at E and F respectively 3.Triangle ABC = triangle AEF 3.lf two angles of one triangle are equal respectively to two angles of another, then the triangle are similar. (a.a.) 4.AE + EB = AB & AF+FC = AC 4.Segment Addition Postulate %3D 5.Triangle BDE = triangle ADE & triangle CDF = triangle ADF 5.Definition of angle bisector %3D 6.Corresponding sides of similar triangles are proportional (C.S.S.T.P.) 6.AE/EB = AF/FC %3D 7.Angle ABD = angle AEF & angle BCA = 7.Corresponding Angles Postulate angle EFA %3D 8.DE bisects AB and DF bisects AC 8. proportionally 9. If a line divides two sides of a triangle proportionally, then it is parallel to the third side. (Theorem 54) 9.EF // BC ET !!

AutoSave Exercises 4_15 proofs 1 and. . Marc Skwarczynski ON Exercise 4.15 #3: A triangle ABC is inscribed in a circle. M is the midpoint of AMČ and BM intersects AC at D. File Home Insert Draw Design Layout References Mailings Review View Help Table Design Layout 1D Exercises 4.15 #3 Given: Inscribed AABC; M midpoint of AMC; BM intersects AC in D AB AD Prove: %3D BC DC AB Prove: BC AD Statements Reasons DC 1. (see above) 1. Given 2. CM = AM 2. Def. of midpoint 3. ZCBM = - CM; 3. An inscribed angle = ½ the intercepted arc. %3D %3D ZABM = AM M. 4. - CM = - AM 4. Division (of step 2) %3D 2. 5. Substitution 5. ZCBM = LABM %3D (step 3 -> step 4) 6. Def. of bisector 6. BM bisects ZABC 7. The bisector of an interior angle of a divides the oppos internally into se which have the sa AB 7. BC AD %3D DC as the other two 257 words fecus