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ABC is a triangle with AB shorter than side AC. The angle bisector of ∠A intersect BC at D. Given that point E is on the median that's drawn from A, so that BE⊥AD, how do I show that DE||AB? I tried to prove similar triangles, but can't find a way to finish.

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  • $\begingroup$ could you attach a diagram? $\endgroup$
    – math635
    Dec 23, 2015 at 4:22

1 Answer 1

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Thanks for the interesting problem. Here is a sketch of the proof.

Mark the middle point of $BC$ as $M$, so $E$ lies on $AM$. Let the intersections of the line $BE$ with $AD$ and $AC$ be $H$ and $K$, respectively. Finally, let us mark the middle point of $KC$ as $N$, and join $MN$. We shall show $DE \parallel AB$ by proving $$ \frac{ME}{MA} = \frac{MD}{MB}. \qquad (1) $$

Figure

First, we can show that $MN \parallel BK$. This is because in $\triangle CBK$, $M$ and $N$ are the middle points of $CB$ and $CK$, respectively. This means $EK \parallel MN$ in $\triangle AMN$, and it follows $$ \frac{ME}{MA} = \frac{NK}{NA} = \frac{CN}{NA}. \qquad (2) $$

Second, we can similarly show that $HN \parallel BC$. This is because in $\triangle KBC$, $H$ and $N$ are the middle points of $KB$ and $KC$, respectively. It follows that, in $\triangle ADC$, $$ \frac{CN}{NA} = \frac{DH}{HA}. \qquad (3) $$

Third, we can show that $HM \parallel CK$. Then in $\triangle DAC$, $$ \frac{DH}{HA} = \frac{DM}{MC} = \frac{MD}{MB}. \qquad (4) $$ Combining (2), (3) and (4) yields (1).

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