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A synthetic geometrical proof required on a result regarding Isosceles triangles.
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I need a synthetic proof on this problem without the use of trigonometry.
Question:
Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AFperp BE$.
The proof is very easy using coordinate geometry.
I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.
geometry euclidean-geometry triangle
asked Aug 1 at 10:37
Love Invariants
77715
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up vote
3
down vote
favorite
I need a synthetic proof on this problem without the use of trigonometry.
Question:
Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AFperp BE$.
The proof is very easy using coordinate geometry.
I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.
geometry euclidean-geometry triangle
asked Aug 1 at 10:37
Love Invariants
77715
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I need a synthetic proof on this problem without the use of trigonometry.
Question:
Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AFperp BE$.
The proof is very easy using coordinate geometry.
I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.
geometry euclidean-geometry triangle
asked Aug 1 at 10:37
Love Invariants
77715
I need a synthetic proof on this problem without the use of trigonometry.
Question:
Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AFperp BE$.
The proof is very easy using coordinate geometry.
I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.
geometry euclidean-geometry triangle
asked Aug 1 at 10:37
Love Invariants
77715
edited Aug 1 at 10:44
asked Aug 1 at 10:37
Love Invariants
77715
asked Aug 1 at 10:37
Love Invariants
77715
asked Aug 1 at 10:37
Love Invariants
77715
77715
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
4
down vote
accepted
Let $G$ be a midpoint for $CE$. Then $DG$ is a middle line in triangle $BCE$ so $DG||BE$. Also we see that $FG||DC$ so $FGbot AD$. So $F$ is an orthocenter in triangle $ADG$. So ...
answered Aug 1 at 11:28
greedoid
26.1k93473
Nice solution! +1
– Michael Rozenberg
Aug 1 at 11:48
Really nice solution.
– Love Invariants
Aug 1 at 16:05
add a comment |Â
up vote
1
down vote
Let $K$ be a midpoint of $BD$.
Thus, since $$Delta ABDsimDelta DCE,$$ we obtain:
$$fracABDC=fracBDCE$$ or
$$frac2ABBC=frac2BKCE$$ or
$$fracABBC=fracBKCE$$ and since $measuredangle ABK=measuredangle BCE,$ we obtain
$$Delta ABKsimDelta BCE,$$
which gives
$$measuredangle BKA=measuredangle BEC.$$
Now, let $BEcap AK=L.$
Thus, $$measuredangle BLA=measuredangle LAE+measuredangle BEA=measuredangle KAD+180^circ-measuredangle BEC=$$
$$=measuredangle KAD+180^circ-measuredangle BKA=180^circ-90^circ=90^circ.$$
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $G$ be a midpoint for $CE$. Then $DG$ is a middle line in triangle $BCE$ so $DG||BE$. Also we see that $FG||DC$ so $FGbot AD$. So $F$ is an orthocenter in triangle $ADG$. So ...
answered Aug 1 at 11:28
greedoid
26.1k93473
Nice solution! +1
– Michael Rozenberg
Aug 1 at 11:48
Really nice solution.
– Love Invariants
Aug 1 at 16:05
add a comment |Â
up vote
4
down vote
accepted
Let $G$ be a midpoint for $CE$. Then $DG$ is a middle line in triangle $BCE$ so $DG||BE$. Also we see that $FG||DC$ so $FGbot AD$. So $F$ is an orthocenter in triangle $ADG$. So ...
answered Aug 1 at 11:28
greedoid
26.1k93473
Nice solution! +1
– Michael Rozenberg
Aug 1 at 11:48
Really nice solution.
– Love Invariants
Aug 1 at 16:05
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $G$ be a midpoint for $CE$. Then $DG$ is a middle line in triangle $BCE$ so $DG||BE$. Also we see that $FG||DC$ so $FGbot AD$. So $F$ is an orthocenter in triangle $ADG$. So ...
answered Aug 1 at 11:28
greedoid
26.1k93473
Let $G$ be a midpoint for $CE$. Then $DG$ is a middle line in triangle $BCE$ so $DG||BE$. Also we see that $FG||DC$ so $FGbot AD$. So $F$ is an orthocenter in triangle $ADG$. So ...
answered Aug 1 at 11:28
greedoid
26.1k93473
answered Aug 1 at 11:28
greedoid
26.1k93473
answered Aug 1 at 11:28
greedoid
26.1k93473
answered Aug 1 at 11:28
greedoid
26.1k93473
26.1k93473
Nice solution! +1
– Michael Rozenberg
Aug 1 at 11:48
Really nice solution.
– Love Invariants
Aug 1 at 16:05
add a comment |Â
Nice solution! +1
– Michael Rozenberg
Aug 1 at 11:48
Really nice solution.
– Love Invariants
Aug 1 at 16:05
Nice solution! +1
– Michael Rozenberg
Aug 1 at 11:48
Nice solution! +1
– Michael Rozenberg
Aug 1 at 11:48
Really nice solution.
– Love Invariants
Aug 1 at 16:05
Really nice solution.
– Love Invariants
Aug 1 at 16:05
add a comment |Â
up vote
1
down vote
Let $K$ be a midpoint of $BD$.
Thus, since $$Delta ABDsimDelta DCE,$$ we obtain:
$$fracABDC=fracBDCE$$ or
$$frac2ABBC=frac2BKCE$$ or
$$fracABBC=fracBKCE$$ and since $measuredangle ABK=measuredangle BCE,$ we obtain
$$Delta ABKsimDelta BCE,$$
which gives
$$measuredangle BKA=measuredangle BEC.$$
Now, let $BEcap AK=L.$
Thus, $$measuredangle BLA=measuredangle LAE+measuredangle BEA=measuredangle KAD+180^circ-measuredangle BEC=$$
$$=measuredangle KAD+180^circ-measuredangle BKA=180^circ-90^circ=90^circ.$$
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
add a comment |Â
up vote
1
down vote
Let $K$ be a midpoint of $BD$.
Thus, since $$Delta ABDsimDelta DCE,$$ we obtain:
$$fracABDC=fracBDCE$$ or
$$frac2ABBC=frac2BKCE$$ or
$$fracABBC=fracBKCE$$ and since $measuredangle ABK=measuredangle BCE,$ we obtain
$$Delta ABKsimDelta BCE,$$
which gives
$$measuredangle BKA=measuredangle BEC.$$
Now, let $BEcap AK=L.$
Thus, $$measuredangle BLA=measuredangle LAE+measuredangle BEA=measuredangle KAD+180^circ-measuredangle BEC=$$
$$=measuredangle KAD+180^circ-measuredangle BKA=180^circ-90^circ=90^circ.$$
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $K$ be a midpoint of $BD$.
Thus, since $$Delta ABDsimDelta DCE,$$ we obtain:
$$fracABDC=fracBDCE$$ or
$$frac2ABBC=frac2BKCE$$ or
$$fracABBC=fracBKCE$$ and since $measuredangle ABK=measuredangle BCE,$ we obtain
$$Delta ABKsimDelta BCE,$$
which gives
$$measuredangle BKA=measuredangle BEC.$$
Now, let $BEcap AK=L.$
Thus, $$measuredangle BLA=measuredangle LAE+measuredangle BEA=measuredangle KAD+180^circ-measuredangle BEC=$$
$$=measuredangle KAD+180^circ-measuredangle BKA=180^circ-90^circ=90^circ.$$
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
Let $K$ be a midpoint of $BD$.
Thus, since $$Delta ABDsimDelta DCE,$$ we obtain:
$$fracABDC=fracBDCE$$ or
$$frac2ABBC=frac2BKCE$$ or
$$fracABBC=fracBKCE$$ and since $measuredangle ABK=measuredangle BCE,$ we obtain
$$Delta ABKsimDelta BCE,$$
which gives
$$measuredangle BKA=measuredangle BEC.$$
Now, let $BEcap AK=L.$
Thus, $$measuredangle BLA=measuredangle LAE+measuredangle BEA=measuredangle KAD+180^circ-measuredangle BEC=$$
$$=measuredangle KAD+180^circ-measuredangle BKA=180^circ-90^circ=90^circ.$$
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
answered Aug 1 at 11:36
Michael Rozenberg
87.4k1577179
87.4k1577179
add a comment |Â
add a comment |Â
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