In the square $ABCD$, prove that $BF+DE=AE$.
$begingroup$
Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$
calculus geometry area
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
$endgroup$
add a comment |
$begingroup$
Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$
calculus geometry area
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
$endgroup$
$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18
$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20
$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26
$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31
$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32
add a comment |
$begingroup$
Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$
calculus geometry area
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
$endgroup$
Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$
calculus geometry area
calculus geometry area
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
edited Nov 23 '18 at 19:49
soodehMehboodi
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
asked Nov 23 '18 at 19:16
soodehMehboodisoodehMehboodi
55428
55428
$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18
$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20
$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26
$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31
$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32
add a comment |
$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18
$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20
$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26
$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31
$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32
$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18
$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18
$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20
$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20
$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26
$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26
$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31
$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31
$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32
$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
$endgroup$
add a comment |
$begingroup$
Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
$$
\BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
\AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
\BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
\cos2alpha=1-2sin^2alpha
$$
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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$begingroup$
Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
$endgroup$
add a comment |
$begingroup$
Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
$endgroup$
add a comment |
$begingroup$
Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
$endgroup$
Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
edited Nov 23 '18 at 20:02
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
answered Nov 23 '18 at 19:53
greedoidgreedoid
38.6k114797
38.6k114797
add a comment |
add a comment |
$begingroup$
Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
$$
\BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
\AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
\BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
\cos2alpha=1-2sin^2alpha
$$
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
$endgroup$
add a comment |
$begingroup$
Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
$$
\BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
\AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
\BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
\cos2alpha=1-2sin^2alpha
$$
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
$endgroup$
add a comment |
$begingroup$
Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
$$
\BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
\AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
\BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
\cos2alpha=1-2sin^2alpha
$$
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
$endgroup$
Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
$$
\BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
\AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
\BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
\cos2alpha=1-2sin^2alpha
$$
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
answered Nov 23 '18 at 19:53
Samvel SafaryanSamvel Safaryan
501111
501111
add a comment |
add a comment |
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$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18
$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20
$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26
$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31
$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32