2 answers:
Answer:
prove it by S.S.S axiom bro
Step-by-step explanation:
statement reason
1.in triangle ADB and ADC
a. AB=AC a.given
b.BD=CD b. radii of same circle
c. AD=AD c. common side
2. ∵triangle ADB is congurent to 2. by S.S.S axiom
ADC
Answer:
AD cuts BC at I => I is the midpoint of BC => BI = CI
ΔAIB and ΔAIC have:
AB = AC
BI = CI
m∠ABC = m∠ACB ( because ΔABC is a isosceles triangle)
=> ΔAIB ≅ ΔAIC (sas)
=> m∠BAD = m∠CAD
ΔADB and ΔADC have:
m∠BAD = m∠CAD
AB = AC
AD is the common edge
=> ΔADB ≅ ΔADC (sas)
Step-by-step explanation:
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Answer:
Im pretty sure the anwer is B.
Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Answer:
x=64°
Step-by-step explanation:
x=180-90-26 (angles on a str line)
=64°
Answer:
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