find the angle in a triangle with angles $ 20^{circ}, 70^circ, 90^circ $
$begingroup$
I have triangle geometry problem:
a) Let $triangle ABC$ be a right triangle with $angle A=90^circ$ and $angle B=20^circ$. Let BE be the angle bisector of $angle B$, and $F$ be a point on segment $AB$ such that $angle ACF=30^circ$. prove that $angle CFE=20^circ$.
b) Prove the same statement with the condition $angle ACF=40^circ$ instead of $angle ACF=30^circ$.
euclidean-geometry triangle angle
edited Jan 21 at 9:58
user549397
1,5081418
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
$endgroup$
add a comment |
$begingroup$
I have triangle geometry problem:
a) Let $triangle ABC$ be a right triangle with $angle A=90^circ$ and $angle B=20^circ$. Let BE be the angle bisector of $angle B$, and $F$ be a point on segment $AB$ such that $angle ACF=30^circ$. prove that $angle CFE=20^circ$.
b) Prove the same statement with the condition $angle ACF=40^circ$ instead of $angle ACF=30^circ$.
euclidean-geometry triangle angle
edited Jan 21 at 9:58
user549397
1,5081418
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
$endgroup$
$begingroup$
Try drawing them out and using the trigonometry formulae that you have been taught.
$endgroup$
– Bill Wallis
May 4 '18 at 12:41
1
$begingroup$
Thanks Mr. Wallis, but I'm searching for a purely geometric and non-trig solution.
$endgroup$
– Saeed_T
May 4 '18 at 13:36
$begingroup$
If only purely geometric means may be used, then it is evident that the techniques @BillWallis suggested are not enough to determine the direction of $EF$ with respect to the other lines' directions, and not enough to solve this problem.
$endgroup$
– Rosie F
Jan 21 at 9:47
add a comment |
$begingroup$
I have triangle geometry problem:
a) Let $triangle ABC$ be a right triangle with $angle A=90^circ$ and $angle B=20^circ$. Let BE be the angle bisector of $angle B$, and $F$ be a point on segment $AB$ such that $angle ACF=30^circ$. prove that $angle CFE=20^circ$.
b) Prove the same statement with the condition $angle ACF=40^circ$ instead of $angle ACF=30^circ$.
euclidean-geometry triangle angle
edited Jan 21 at 9:58
user549397
1,5081418
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
$endgroup$
I have triangle geometry problem:
a) Let $triangle ABC$ be a right triangle with $angle A=90^circ$ and $angle B=20^circ$. Let BE be the angle bisector of $angle B$, and $F$ be a point on segment $AB$ such that $angle ACF=30^circ$. prove that $angle CFE=20^circ$.
b) Prove the same statement with the condition $angle ACF=40^circ$ instead of $angle ACF=30^circ$.
euclidean-geometry triangle angle
euclidean-geometry triangle angle
edited Jan 21 at 9:58
user549397
1,5081418
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
edited Jan 21 at 9:58
user549397
1,5081418
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
edited Jan 21 at 9:58
user549397
1,5081418
edited Jan 21 at 9:58
user549397
1,5081418
edited Jan 21 at 9:58
user549397
1,5081418
1,5081418
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
asked May 4 '18 at 6:40
Saeed_TSaeed_T
575
575
$begingroup$
Try drawing them out and using the trigonometry formulae that you have been taught.
$endgroup$
– Bill Wallis
May 4 '18 at 12:41
1
$begingroup$
Thanks Mr. Wallis, but I'm searching for a purely geometric and non-trig solution.
$endgroup$
– Saeed_T
May 4 '18 at 13:36
$begingroup$
If only purely geometric means may be used, then it is evident that the techniques @BillWallis suggested are not enough to determine the direction of $EF$ with respect to the other lines' directions, and not enough to solve this problem.
$endgroup$
– Rosie F
Jan 21 at 9:47
add a comment |
$begingroup$
Try drawing them out and using the trigonometry formulae that you have been taught.
$endgroup$
– Bill Wallis
May 4 '18 at 12:41
1
$begingroup$
Thanks Mr. Wallis, but I'm searching for a purely geometric and non-trig solution.
$endgroup$
– Saeed_T
May 4 '18 at 13:36
$begingroup$
If only purely geometric means may be used, then it is evident that the techniques @BillWallis suggested are not enough to determine the direction of $EF$ with respect to the other lines' directions, and not enough to solve this problem.
$endgroup$
– Rosie F
Jan 21 at 9:47
$begingroup$
Try drawing them out and using the trigonometry formulae that you have been taught.
$endgroup$
– Bill Wallis
May 4 '18 at 12:41
$begingroup$
Try drawing them out and using the trigonometry formulae that you have been taught.
$endgroup$
– Bill Wallis
May 4 '18 at 12:41
1
1
$begingroup$
Thanks Mr. Wallis, but I'm searching for a purely geometric and non-trig solution.
$endgroup$
– Saeed_T
May 4 '18 at 13:36
$begingroup$
Thanks Mr. Wallis, but I'm searching for a purely geometric and non-trig solution.
$endgroup$
– Saeed_T
May 4 '18 at 13:36
$begingroup$
If only purely geometric means may be used, then it is evident that the techniques @BillWallis suggested are not enough to determine the direction of $EF$ with respect to the other lines' directions, and not enough to solve this problem.
$endgroup$
– Rosie F
Jan 21 at 9:47
$begingroup$
If only purely geometric means may be used, then it is evident that the techniques @BillWallis suggested are not enough to determine the direction of $EF$ with respect to the other lines' directions, and not enough to solve this problem.
$endgroup$
– Rosie F
Jan 21 at 9:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a solution to part (b).
Let $Omega$ be $BCE$'s circumcircle and $O$ its centre. Then $angle COE=2angle CBE=20^circ$. $angle EOB=2angle ECB=140^circ$ so $angle COB=angle COE+angle EOB=160^circ$ so $angle OBC=angle BCO=10^circ$.
Erect an equilateral $triangle BOJ$ on base $BO$.
$OJ=OB$ so $J$ is on $Omega$. $angle JCB=frac{1}2 angle JOB=30^circ=angle FCB$ so $F$ is on $CJ$.
Let $OJ$ cross $BF$ at $P$. $angle OBJ=60^circ=2angle OBF$ so $BF$ is an axis of symmetry of $triangle BOJ$ and is thus $OJ$'s $perp$ bisector. Thus $angle FOJ=angle OJF$.
$OJ=OB=OC$ so $triangle CJO$ is isosceles on base $CJ$. $angle JCO=angle FCO=angle FCB+angle BCO=40^circ$, so $angle OJC=angle JCO=40^circ$ so $angle FOJ=angle OJF=angle OJC=40^circ$.
Therefore $angle EOF=angle EOB-angle FOJ-angle JOB=140^circ-40^circ-60^circ=40^circ=angle ECF$, so $OCEF$ is cyclic, so $angle CFE=angle COE=20^circ$, which solves part b.
Hints for part (a): it might be useful to know that $OCEF_aF_b$ are all on a circle whose centre lies on $BC$.
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
$endgroup$
add a comment |
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$begingroup$
Here is a solution to part (b).
Let $Omega$ be $BCE$'s circumcircle and $O$ its centre. Then $angle COE=2angle CBE=20^circ$. $angle EOB=2angle ECB=140^circ$ so $angle COB=angle COE+angle EOB=160^circ$ so $angle OBC=angle BCO=10^circ$.
Erect an equilateral $triangle BOJ$ on base $BO$.
$OJ=OB$ so $J$ is on $Omega$. $angle JCB=frac{1}2 angle JOB=30^circ=angle FCB$ so $F$ is on $CJ$.
Let $OJ$ cross $BF$ at $P$. $angle OBJ=60^circ=2angle OBF$ so $BF$ is an axis of symmetry of $triangle BOJ$ and is thus $OJ$'s $perp$ bisector. Thus $angle FOJ=angle OJF$.
$OJ=OB=OC$ so $triangle CJO$ is isosceles on base $CJ$. $angle JCO=angle FCO=angle FCB+angle BCO=40^circ$, so $angle OJC=angle JCO=40^circ$ so $angle FOJ=angle OJF=angle OJC=40^circ$.
Therefore $angle EOF=angle EOB-angle FOJ-angle JOB=140^circ-40^circ-60^circ=40^circ=angle ECF$, so $OCEF$ is cyclic, so $angle CFE=angle COE=20^circ$, which solves part b.
Hints for part (a): it might be useful to know that $OCEF_aF_b$ are all on a circle whose centre lies on $BC$.
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
$endgroup$
add a comment |
$begingroup$
Here is a solution to part (b).
Let $Omega$ be $BCE$'s circumcircle and $O$ its centre. Then $angle COE=2angle CBE=20^circ$. $angle EOB=2angle ECB=140^circ$ so $angle COB=angle COE+angle EOB=160^circ$ so $angle OBC=angle BCO=10^circ$.
Erect an equilateral $triangle BOJ$ on base $BO$.
$OJ=OB$ so $J$ is on $Omega$. $angle JCB=frac{1}2 angle JOB=30^circ=angle FCB$ so $F$ is on $CJ$.
Let $OJ$ cross $BF$ at $P$. $angle OBJ=60^circ=2angle OBF$ so $BF$ is an axis of symmetry of $triangle BOJ$ and is thus $OJ$'s $perp$ bisector. Thus $angle FOJ=angle OJF$.
$OJ=OB=OC$ so $triangle CJO$ is isosceles on base $CJ$. $angle JCO=angle FCO=angle FCB+angle BCO=40^circ$, so $angle OJC=angle JCO=40^circ$ so $angle FOJ=angle OJF=angle OJC=40^circ$.
Therefore $angle EOF=angle EOB-angle FOJ-angle JOB=140^circ-40^circ-60^circ=40^circ=angle ECF$, so $OCEF$ is cyclic, so $angle CFE=angle COE=20^circ$, which solves part b.
Hints for part (a): it might be useful to know that $OCEF_aF_b$ are all on a circle whose centre lies on $BC$.
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
$endgroup$
add a comment |
$begingroup$
Here is a solution to part (b).
Let $Omega$ be $BCE$'s circumcircle and $O$ its centre. Then $angle COE=2angle CBE=20^circ$. $angle EOB=2angle ECB=140^circ$ so $angle COB=angle COE+angle EOB=160^circ$ so $angle OBC=angle BCO=10^circ$.
Erect an equilateral $triangle BOJ$ on base $BO$.
$OJ=OB$ so $J$ is on $Omega$. $angle JCB=frac{1}2 angle JOB=30^circ=angle FCB$ so $F$ is on $CJ$.
Let $OJ$ cross $BF$ at $P$. $angle OBJ=60^circ=2angle OBF$ so $BF$ is an axis of symmetry of $triangle BOJ$ and is thus $OJ$'s $perp$ bisector. Thus $angle FOJ=angle OJF$.
$OJ=OB=OC$ so $triangle CJO$ is isosceles on base $CJ$. $angle JCO=angle FCO=angle FCB+angle BCO=40^circ$, so $angle OJC=angle JCO=40^circ$ so $angle FOJ=angle OJF=angle OJC=40^circ$.
Therefore $angle EOF=angle EOB-angle FOJ-angle JOB=140^circ-40^circ-60^circ=40^circ=angle ECF$, so $OCEF$ is cyclic, so $angle CFE=angle COE=20^circ$, which solves part b.
Hints for part (a): it might be useful to know that $OCEF_aF_b$ are all on a circle whose centre lies on $BC$.
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
$endgroup$
Here is a solution to part (b).
Let $Omega$ be $BCE$'s circumcircle and $O$ its centre. Then $angle COE=2angle CBE=20^circ$. $angle EOB=2angle ECB=140^circ$ so $angle COB=angle COE+angle EOB=160^circ$ so $angle OBC=angle BCO=10^circ$.
Erect an equilateral $triangle BOJ$ on base $BO$.
$OJ=OB$ so $J$ is on $Omega$. $angle JCB=frac{1}2 angle JOB=30^circ=angle FCB$ so $F$ is on $CJ$.
Let $OJ$ cross $BF$ at $P$. $angle OBJ=60^circ=2angle OBF$ so $BF$ is an axis of symmetry of $triangle BOJ$ and is thus $OJ$'s $perp$ bisector. Thus $angle FOJ=angle OJF$.
$OJ=OB=OC$ so $triangle CJO$ is isosceles on base $CJ$. $angle JCO=angle FCO=angle FCB+angle BCO=40^circ$, so $angle OJC=angle JCO=40^circ$ so $angle FOJ=angle OJF=angle OJC=40^circ$.
Therefore $angle EOF=angle EOB-angle FOJ-angle JOB=140^circ-40^circ-60^circ=40^circ=angle ECF$, so $OCEF$ is cyclic, so $angle CFE=angle COE=20^circ$, which solves part b.
Hints for part (a): it might be useful to know that $OCEF_aF_b$ are all on a circle whose centre lies on $BC$.
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
answered Jan 21 at 9:43
Rosie FRosie F
1,312315
1,312315
add a comment |
add a comment |
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$begingroup$
Try drawing them out and using the trigonometry formulae that you have been taught.
$endgroup$
– Bill Wallis
May 4 '18 at 12:41
1
$begingroup$
Thanks Mr. Wallis, but I'm searching for a purely geometric and non-trig solution.
$endgroup$
– Saeed_T
May 4 '18 at 13:36
$begingroup$
If only purely geometric means may be used, then it is evident that the techniques @BillWallis suggested are not enough to determine the direction of $EF$ with respect to the other lines' directions, and not enough to solve this problem.
$endgroup$
– Rosie F
Jan 21 at 9:47