Mixing
What is the length of $EF$ in the following diagram?
$begingroup$
In triangle $triangle ABC$, angle $A=50^circ$ , angle $C=65^circ$ .
Point $F$ is on $AC$ such that, $BF$ is perpendicular to $A$C. $D$ is
a point on $BF$ (extended) such that $AD=AB$. E is a point $CD$ such
that, AE is perpendicular to $CD$. If $BC=12$, what is the length of
$EF$?
Source: Bangladesh Math Olympiad 2016 Junior Catagory
I tried and proved that $ABF cong AFD$ and $BCF cong CFD$. I am not able to find any relation of $EF$ with other sides.
geometry contest-math triangle congruences-geometry
asked Jan 26 at 13:34
ShromiShromi
3289
$endgroup$
add a comment |
$begingroup$
In triangle $triangle ABC$, angle $A=50^circ$ , angle $C=65^circ$ .
Point $F$ is on $AC$ such that, $BF$ is perpendicular to $A$C. $D$ is
a point on $BF$ (extended) such that $AD=AB$. E is a point $CD$ such
that, AE is perpendicular to $CD$. If $BC=12$, what is the length of
$EF$?
Source: Bangladesh Math Olympiad 2016 Junior Catagory
I tried and proved that $ABF cong AFD$ and $BCF cong CFD$. I am not able to find any relation of $EF$ with other sides.
geometry contest-math triangle congruences-geometry
asked Jan 26 at 13:34
ShromiShromi
3289
$endgroup$
1
$begingroup$
$F$ is the midpoint of $BD$ and $E$ is the midpoint of $CD$. Hence $EF={1over2}BC$.
$endgroup$
– Aretino
Jan 26 at 14:06
add a comment |
$begingroup$
In triangle $triangle ABC$, angle $A=50^circ$ , angle $C=65^circ$ .
Point $F$ is on $AC$ such that, $BF$ is perpendicular to $A$C. $D$ is
a point on $BF$ (extended) such that $AD=AB$. E is a point $CD$ such
that, AE is perpendicular to $CD$. If $BC=12$, what is the length of
$EF$?
Source: Bangladesh Math Olympiad 2016 Junior Catagory
I tried and proved that $ABF cong AFD$ and $BCF cong CFD$. I am not able to find any relation of $EF$ with other sides.
geometry contest-math triangle congruences-geometry
asked Jan 26 at 13:34
ShromiShromi
3289
$endgroup$
In triangle $triangle ABC$, angle $A=50^circ$ , angle $C=65^circ$ .
Point $F$ is on $AC$ such that, $BF$ is perpendicular to $A$C. $D$ is
a point on $BF$ (extended) such that $AD=AB$. E is a point $CD$ such
that, AE is perpendicular to $CD$. If $BC=12$, what is the length of
$EF$?
Source: Bangladesh Math Olympiad 2016 Junior Catagory
I tried and proved that $ABF cong AFD$ and $BCF cong CFD$. I am not able to find any relation of $EF$ with other sides.
geometry contest-math triangle congruences-geometry
geometry contest-math triangle congruences-geometry
asked Jan 26 at 13:34
ShromiShromi
3289
asked Jan 26 at 13:34
ShromiShromi
3289
asked Jan 26 at 13:34
ShromiShromi
3289
asked Jan 26 at 13:34
ShromiShromi
3289
asked Jan 26 at 13:34
ShromiShromi
3289
3289
1
$begingroup$
$F$ is the midpoint of $BD$ and $E$ is the midpoint of $CD$. Hence $EF={1over2}BC$.
$endgroup$
– Aretino
Jan 26 at 14:06
add a comment |
1
$begingroup$
$F$ is the midpoint of $BD$ and $E$ is the midpoint of $CD$. Hence $EF={1over2}BC$.
$endgroup$
– Aretino
Jan 26 at 14:06
1
1
$begingroup$
$F$ is the midpoint of $BD$ and $E$ is the midpoint of $CD$. Hence $EF={1over2}BC$.
$endgroup$
– Aretino
Jan 26 at 14:06
$begingroup$
$F$ is the midpoint of $BD$ and $E$ is the midpoint of $CD$. Hence $EF={1over2}BC$.
$endgroup$
– Aretino
Jan 26 at 14:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
From the congruent triangles you have already found, you can conclude that
$triangle ABC cong triangle ACD.$
Use the two known angles of $triangle ABC$ to find the third angle.
You will then be able to show that $triangle ABC$ and $triangle ACD$ are isosceles triangles, and that $E$ is the midpoint of $CD.$
Since $AB = AD,$ triangle $triangle ABD$ also is isosceles and $F$ is the midpoint of $BD.$
These facts should tell you something about the relationship between $triangle BCD$ and $triangle FED,$
after which you can find the answer.
answered Jan 26 at 14:01
David KDavid K
55.3k344120
$endgroup$
$begingroup$
Thank you for the answer. I successfully found that EF = 6. That was a great explanation!
$endgroup$
– Shromi
Jan 26 at 15:10
add a comment |
$begingroup$
Angle $CBA=180°-50°-65°=65°=angle ACB$, so $AB=AC=AD$. In triangle $BCF$, $CE=BC/2$, $FC=BCcos 65°$ and $angle FCE = 65°$, so one can find $FE$ with cosine theorem.
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
From the congruent triangles you have already found, you can conclude that
$triangle ABC cong triangle ACD.$
Use the two known angles of $triangle ABC$ to find the third angle.
You will then be able to show that $triangle ABC$ and $triangle ACD$ are isosceles triangles, and that $E$ is the midpoint of $CD.$
Since $AB = AD,$ triangle $triangle ABD$ also is isosceles and $F$ is the midpoint of $BD.$
These facts should tell you something about the relationship between $triangle BCD$ and $triangle FED,$
after which you can find the answer.
answered Jan 26 at 14:01
David KDavid K
55.3k344120
$endgroup$
$begingroup$
Thank you for the answer. I successfully found that EF = 6. That was a great explanation!
$endgroup$
– Shromi
Jan 26 at 15:10
add a comment |
$begingroup$
From the congruent triangles you have already found, you can conclude that
$triangle ABC cong triangle ACD.$
Use the two known angles of $triangle ABC$ to find the third angle.
You will then be able to show that $triangle ABC$ and $triangle ACD$ are isosceles triangles, and that $E$ is the midpoint of $CD.$
Since $AB = AD,$ triangle $triangle ABD$ also is isosceles and $F$ is the midpoint of $BD.$
These facts should tell you something about the relationship between $triangle BCD$ and $triangle FED,$
after which you can find the answer.
answered Jan 26 at 14:01
David KDavid K
55.3k344120
$endgroup$
$begingroup$
Thank you for the answer. I successfully found that EF = 6. That was a great explanation!
$endgroup$
– Shromi
Jan 26 at 15:10
add a comment |
$begingroup$
From the congruent triangles you have already found, you can conclude that
$triangle ABC cong triangle ACD.$
Use the two known angles of $triangle ABC$ to find the third angle.
You will then be able to show that $triangle ABC$ and $triangle ACD$ are isosceles triangles, and that $E$ is the midpoint of $CD.$
Since $AB = AD,$ triangle $triangle ABD$ also is isosceles and $F$ is the midpoint of $BD.$
These facts should tell you something about the relationship between $triangle BCD$ and $triangle FED,$
after which you can find the answer.
answered Jan 26 at 14:01
David KDavid K
55.3k344120
$endgroup$
From the congruent triangles you have already found, you can conclude that
$triangle ABC cong triangle ACD.$
Use the two known angles of $triangle ABC$ to find the third angle.
You will then be able to show that $triangle ABC$ and $triangle ACD$ are isosceles triangles, and that $E$ is the midpoint of $CD.$
Since $AB = AD,$ triangle $triangle ABD$ also is isosceles and $F$ is the midpoint of $BD.$
These facts should tell you something about the relationship between $triangle BCD$ and $triangle FED,$
after which you can find the answer.
answered Jan 26 at 14:01
David KDavid K
55.3k344120
answered Jan 26 at 14:01
David KDavid K
55.3k344120
answered Jan 26 at 14:01
David KDavid K
55.3k344120
answered Jan 26 at 14:01
David KDavid K
55.3k344120
55.3k344120
$begingroup$
Thank you for the answer. I successfully found that EF = 6. That was a great explanation!
$endgroup$
– Shromi
Jan 26 at 15:10
add a comment |
$begingroup$
Thank you for the answer. I successfully found that EF = 6. That was a great explanation!
$endgroup$
– Shromi
Jan 26 at 15:10
$begingroup$
Thank you for the answer. I successfully found that EF = 6. That was a great explanation!
$endgroup$
– Shromi
Jan 26 at 15:10
$begingroup$
Thank you for the answer. I successfully found that EF = 6. That was a great explanation!
$endgroup$
– Shromi
Jan 26 at 15:10
add a comment |
$begingroup$
Angle $CBA=180°-50°-65°=65°=angle ACB$, so $AB=AC=AD$. In triangle $BCF$, $CE=BC/2$, $FC=BCcos 65°$ and $angle FCE = 65°$, so one can find $FE$ with cosine theorem.
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
Angle $CBA=180°-50°-65°=65°=angle ACB$, so $AB=AC=AD$. In triangle $BCF$, $CE=BC/2$, $FC=BCcos 65°$ and $angle FCE = 65°$, so one can find $FE$ with cosine theorem.
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
Angle $CBA=180°-50°-65°=65°=angle ACB$, so $AB=AC=AD$. In triangle $BCF$, $CE=BC/2$, $FC=BCcos 65°$ and $angle FCE = 65°$, so one can find $FE$ with cosine theorem.
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
$endgroup$
Angle $CBA=180°-50°-65°=65°=angle ACB$, so $AB=AC=AD$. In triangle $BCF$, $CE=BC/2$, $FC=BCcos 65°$ and $angle FCE = 65°$, so one can find $FE$ with cosine theorem.
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
answered Jan 26 at 13:39
Vasily MitchVasily Mitch
2,6791312
2,6791312
add a comment |
add a comment |
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1
$begingroup$
$F$ is the midpoint of $BD$ and $E$ is the midpoint of $CD$. Hence $EF={1over2}BC$.
$endgroup$
– Aretino
Jan 26 at 14:06