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Ivahew [28]
3 years ago
10

Prove that a line parallel to one side of a triangle divides the other two sides proportionally. Be sure to create and name the

appropriate geometric figures. (10 points)
WILL GIVE BRAINLIEST IF CORRECT

Mathematics
1 answer:
ryzh [129]3 years ago
7 0

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)

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katovenus [111]
   
\displaystyle\\
1 \frac{2}{5} +\Big(-5 \frac{1}{2} \Big) = ? \\  \\ 
1 \frac{2}{5} = \frac{1\times 5+2}{5}=\boxed{\frac{7}{5}} \\  \\ 
-5 \frac{1}{2} =-5 -\frac{1}{2} = \frac{-5\times2-1}{2} =\frac{-10-1}{2}=\frac{-11}{2}= \boxed{-\frac{11}{2} }\\  \\ \texttt{OR} \\  \\ 
-5 \frac{1}{2} = -\Big(5 \frac{1}{2} \Big)= -\Big( \frac{5\times2+1}{2} \Big)=-\Big( \frac{11}{2} \Big)= \boxed{-\frac{11}{2} }


\displaystyle\\
\Longrightarrow ~~1 \frac{2}{5} +\Big(-5 \frac{1}{2} \Big) =\frac{7}{5} -\frac{11}{2} = \frac{7\times 2}{5\times 2} -\frac{11\times 5}{2\times 5} = \\  \\ 
= \frac{14}{10} -\frac{55}{10} = \frac{14-55}{10} =\frac{-41}{10} = -\frac{41}{10}=-\frac{40+1}{10}=\boxed{\boxed{-4\frac{1}{10}}}



7 0
3 years ago
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