/5. D is the midpoint of side BC of triangle ABC and the bisectors of angles ADB and ADC meet AB and AC at E and F respectively. Prove: EF is parallel to BC. (See Theo- rem 54.) 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

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter8: Areas Of Polygons And Circles
Section8.5: More Area Relationships In The Circle
Problem 9E
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Theorem 54=If a line divides two sides of a triangle proportionally, then it is parallel to the third side. 

Please answer in a two-column proof-statements and reasons. **Also, it should have ONLY one given (statement).** I have an example of how it should look like/how it should be done in the second picture.

It is about lines proportional. The question is in the picture. *The question is completely clear, so please don't reject the question.*

/5. D is the midpoint of side BC of triangle
ABC and the bisectors of angles ADB
and ADC meet AB and AC at E and F
respectively.
Prove: EF is parallel to BC. (See Theo-
rem 54.)
Transcribed Image Text:/5. D is the midpoint of side BC of triangle ABC and the bisectors of angles ADB and ADC meet AB and AC at E and F respectively. Prove: EF is parallel to BC. (See Theo- rem 54.)
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
Transcribed Image Text: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
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