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1) In the diagram below of parallelogram ABCD, AFGB, CF bisects ZDCB, DG bisects ZADC, and CF and DG intersect at E. E A F G B If m/B= 75°, then the measure of ZEFA is 1) 142.5° 2) 127.5⁰ 3) 52.5° 4) 37.5° I

1) In the diagram below of parallelogram ABCD, AFGB, CF bisects ZDCB, DG bisects ZADC, and CF and DG intersect at E. E A F G B If m/B= 75°, then the measure of ZEFA is 1) 142.5° 2) 127.5⁰ 3) 52.5° 4) 37.5° I

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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### Geometry: Parallelogram and Angle Bisectors #### Diagram Analysis and Question In the diagram below of parallelogram ABCD: - Line AFGB is drawn. - Line CF bisects ∠DCB. - Line DG bisects ∠ADC. - Lines CF and DG intersect at point E. Below is the diagram as described: ``` D-------------C \ / E /\ F \ / \ / A--------B F------G ``` Given the information: 1. If \( m∠D = 75° \), then the measure of ∠EFA is: 1. 142.5° 2. 127.5° 3. 52.5° 4. 37.5° #### Steps to Solve: 1. Since \( m∠D = 75° \) is given for parallelogram ABCD, we understand from the properties of a parallelogram (opposite angles are equal and adjacent angles sum up to 180°), that: - \( m∠ADC = 75° \) - \( m∠DCB = 105° (since 180 - 75 = 105) \) 2. CF bisects ∠DCB (which means CF splits ∠DCB into two equal angles, each of \( \frac{105°}{2} = 52.5° \)). 3. DG bisects ∠ADC (which means DG splits ∠ADC into two equal angles, each of \( \frac{75°}{2} = 37.5° \)). 4. To find ∠EFA: - Since CF and DG are bisectors intersecting at E, ∠EFA will be equal to the bisected angle by DG direct from point F to E to A. Therefore, \(\angle EFA\) is \( 37.5°\). ### Answer: 4) 37.5° ### Detailed Diagram Explanation: The diagram represents a parallelogram (ABCD), with multiple lines intersecting and bisecting angles at specific points. The two bisector lines (CF and DG) intersect at point E, forming angles at various sections in the shape. The problem focuses on understanding the property of the parallelogram and the way bisectors create equal angles from a
Transcribed Image Text:### Geometry: Parallelogram and Angle Bisectors #### Diagram Analysis and Question In the diagram below of parallelogram ABCD: - Line AFGB is drawn. - Line CF bisects ∠DCB. - Line DG bisects ∠ADC. - Lines CF and DG intersect at point E. Below is the diagram as described: ``` D-------------C \ / E /\ F \ / \ / A--------B F------G ``` Given the information: 1. If \( m∠D = 75° \), then the measure of ∠EFA is: 1. 142.5° 2. 127.5° 3. 52.5° 4. 37.5° #### Steps to Solve: 1. Since \( m∠D = 75° \) is given for parallelogram ABCD, we understand from the properties of a parallelogram (opposite angles are equal and adjacent angles sum up to 180°), that: - \( m∠ADC = 75° \) - \( m∠DCB = 105° (since 180 - 75 = 105) \) 2. CF bisects ∠DCB (which means CF splits ∠DCB into two equal angles, each of \( \frac{105°}{2} = 52.5° \)). 3. DG bisects ∠ADC (which means DG splits ∠ADC into two equal angles, each of \( \frac{75°}{2} = 37.5° \)). 4. To find ∠EFA: - Since CF and DG are bisectors intersecting at E, ∠EFA will be equal to the bisected angle by DG direct from point F to E to A. Therefore, \(\angle EFA\) is \( 37.5°\). ### Answer: 4) 37.5° ### Detailed Diagram Explanation: The diagram represents a parallelogram (ABCD), with multiple lines intersecting and bisecting angles at specific points. The two bisector lines (CF and DG) intersect at point E, forming angles at various sections in the shape. The problem focuses on understanding the property of the parallelogram and the way bisectors create equal angles from a
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