Answer:
in steps
Step-by-step explanation:
DE // BC
m∠ADE = m∠ABC and m∠AED = m∠ACB
∴ ΔADE similar to ΔABC
AB/AD = AC/AE
(AD + DB) / AD = (AE + EC) / AE
AD/AD + DB/AD = AE/AE + EC/AE
1 + DB/AD = 1 + EC/AE
DB/AD = EC/AE (AD/DB = AE/EC)
This is a test i'm taking there's 8 questions. This is due today it would be great if I got help on all the questions
1 answer:
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