Triangle $ABC$ has median $AF$, and $D$ is the midpoint of the median. Line $CD$ intersects $AB$ in $E$. Prove that $BD = BF$ implies $AE = DE$!
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Tags: geometry
thugzmath10
23.07.2016 20:43
We have $BF=CF=BD, FD = DA$ and $\angle CFD=\angle BDA$, so triangles $CFD$ and $BDA$ are congruent. Thus, $\angle CDF=\angle BAD$. But $\angle CDF=\angle ADE$, so $\angle BAD=\angle ADE$. Hence, triangle $ADE$ is isosceles and $AE=DE$.
spacewalker
24.07.2016 09:19
Interesting... @thugzmath10, how did you arrive at the congruency of angles CFD and BDA?
thugzmath10
24.07.2016 16:14
Since $BD=BF$, triangle $BDF$ is isosceles so $\angle BFA = \angle BDF$. It follows that $\angle CFD=180^\circ-\angle BFA=180^\circ-\angle BDF=\angle BDA$.