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 Circumcircles, Incircles, Medians, and Altitudes

I tried to use the angle property by which AD=4 and DB=5,but since F is not given as mid point I don't know how to proceed to find length of DG.I think AED as 90 degree is important but I am unable to figure out its use.

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angle FCE=$C/2$ also FEC=$C/2$ implies FC=EF and also FAE=FEA=$90-C/2$ So EF=AF implies F is the midpoint.

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  • $\begingroup$ Sorry I didn't get how FEC=C/2? $\endgroup$ – swapedoc Jun 23 '14 at 9:16
  • $\begingroup$ AFE=FEC+FCE but AFE=C $\endgroup$ – happymath Jun 23 '14 at 9:18
  • $\begingroup$ Oh yes and also BEC and FEC are alternate angles $\endgroup$ – swapedoc Jun 23 '14 at 9:19
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Referring to the figure (with added constructions) below.

enter image description here

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CONSTRUCTION : extend $AE$ to $BC$ where it intersects $BC$ at $X$.

consider $\triangle AEC$ and $\triangle XCE$.

$\triangle AEC$ and $\triangle XCE$ are congruent by $ASA$.

therefore by CSCTC, $AE = EX$

therefore $E$ is the mid-point of $AX$.

given: $EF$ and $EG$ are parallel to $BC$

therefore by converse of mid-point theorem in $\triangle ACX$ and in $\triangle ABX$, $F$ is the mid-point of $AC$ and $G$ is the mid-point of $AB$.

therefore, $AG = BG = 4.5$

since, you found out $AD = 4$.

therefore,

$DG = AG - AD = 0.5$

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