In Delta ABC,overline{AD} and overline{BE} ar | vedclass

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(B) In $\Delta ABC$,$\overline{BE}$ is a median to side $\overline{AC}$.Therefore,$E$ is the midpoint of $\overline{AC}$,which implies $AE = EC$,or $A

Vedclass · Accepted Answer

$10$

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(B) In $\Delta ABC$,$\overline{BE}$ is a median to side $\overline{AC}$.Therefore,$E$ is the midpoint of $\overline{AC}$,which implies $AE = EC$,or $AC = 2EC$.In $\Delta CEB$,$D$ is the midpoint of $\overline{BC}$ (since $\overline{AD}$ is a median) and line $m$ (passing through $D$) is parallel to $\overline{BE}$.By the Converse of the Midpoint Theorem in $\Delta CEB$,since $D$ is the midpoint of $\overline{BC}$ and $\overline{DK} \parallel \overline{BE}$,$K$ must be the midpoint of $\overline{EC}$.Therefore,$EK = KC = 2.5$.Thus,$EC = EK + KC = 2.5 + 2.5 = 5$.Finally,$AC = 2EC = 2 	imes 5 = 10$.

In $\Delta ABC$,$\overline{AD}$ and $\overline{BE}$ are medians. Line $m$ through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. If $EK = 2.5$,then $AC = \ldots$

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