2. Given: RD and AD are tangent to O G at R and A, respectively, G' E and intersect at the exterior point S Prove: M/RDA = (MREA – mRA) STATEMENTS REASONS 1. Given 1. RD and AD are tangent to 0G at R and A, respectively, and intersect at the exterior point S 2. Two points determine a line 3. MLARM =-MREA and MRAD = 3. %3D 4. MLARM = m/RAD + m2D 4. 5. 5. Substitution 6. MLRDA =;(MREA – mRA) 6. %3D 2.
2. Given: RD and AD are tangent to O G at R and A, respectively, G' E and intersect at the exterior point S Prove: M/RDA = (MREA – mRA) STATEMENTS REASONS 1. Given 1. RD and AD are tangent to 0G at R and A, respectively, and intersect at the exterior point S 2. Two points determine a line 3. MLARM =-MREA and MRAD = 3. %3D 4. MLARM = m/RAD + m2D 4. 5. 5. Substitution 6. MLRDA =;(MREA – mRA) 6. %3D 2.
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter6: Circles
Section6.CT: Test
Problem 6CT: Given the tangents with mRT=146, find : a mRST b m3
Related questions
Question
![M
2. Given: RD and AD are tangent to 0
G
at R and A, respectively,
G'
E
and
intersect at the exterior
A
point S
Prove: m<RDA = ÷(MREA – mRA)
STATEMENTS
REASONS
1. Given
1. RD and AD are tangent to OG
at R and A, respectively, and
intersect at the exterior point S
2.
2. Two points determine a line
3. MLARM =-MREA and MLRAD =
4. MLARM = MLRAD + mLD
4.
5.
5. Substitution
6. MLRDA =÷(MREĀ – mRA)
6.
3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F27938926-7ea3-4559-9237-063f91c56533%2F18685e29-cbdb-487f-bd18-0809188f3bad%2Faiz594f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:M
2. Given: RD and AD are tangent to 0
G
at R and A, respectively,
G'
E
and
intersect at the exterior
A
point S
Prove: m<RDA = ÷(MREA – mRA)
STATEMENTS
REASONS
1. Given
1. RD and AD are tangent to OG
at R and A, respectively, and
intersect at the exterior point S
2.
2. Two points determine a line
3. MLARM =-MREA and MLRAD =
4. MLARM = MLRAD + mLD
4.
5.
5. Substitution
6. MLRDA =÷(MREĀ – mRA)
6.
3.
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