given: △abc,
d is the midpoint
of (overline{ab}) and
e is the midpoint
of (overline{ac}).
prove: (overline{de}paralleloverline{bc}) and
(de=\frac{1}{2}bc).

Question

Accepted Answer

# Explanation:
## Step1: Use the midpoint definition
Since $D$ is the mid - point of $AB$, $AD = DB$. Since $E$ is the mid - point of $AC$, $AE=EC$.
## Step2: Prove $\triangle ADE\cong\triangle CFE$
$\angle AED=\angle CEF$ (vertical angles). $AE = EC$ (mid - point definition). $\angle A=\angle ECF$ (alternate interior angles as $DF\parallel BC$). By $ASA$ (Angle - Side - Angle) congruence criterion, $\triangle ADE\cong\triangle CFE$. Then $AD = CF$ and $DE=EF$. But $AD = DB$, so $DB = CF$.
## Step3: Show that $DBCF$ is a parallelogram
Since $DB\parallel CF$ (because $DF\parallel BC$) and $DB = CF$, by the definition of a parallelogram (a quadrilateral with one pair of opposite sides parallel and equal), $DBCF$ is a parallelogram.
## Step4: Conclude about $DE$ and $BC$
Since $DBCF$ is a parallelogram, $DF = BC$ and $DF\parallel BC$. Also, $DE=\frac{1}{2}DF$ (because $E$ is the mid - point of $DF$ as $DE = EF$). So $DE=\frac{1}{2}BC$ and $DE\parallel BC$ (since $DF\parallel BC$ and $DE$ is a part of $DF$).

# Answer:
15. $AD = DB$ and $AE = EC$
16. $\triangle ADE\cong\triangle CFE$ (by $ASA$) and $DBCF$ is a parallelogram (one pair of opposite sides $DB$ and $CF$ are parallel and equal)

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QUESTION IMAGE

Question

Explanation:

Step1: Use the midpoint definition

Step2: Prove \(\triangle ADE\cong\triangle CFE\)

Step3: Show that \(DBCF\) is a parallelogram

Step4: Conclude about \(DE\) and \(BC\)

Answer: