Right triangle geometry problem
$begingroup$
Right triangle $Delta ABC$ ($angle ACB=90°$). The following is constructed: from point $C$ altitude $CD$, angle bisector $CL$ of $angle ACB$, angle bisector $DK$ of $angle ADC$, angle bisector $DN$ of $angle BDC$.
$D, L$ lie on $AB$, $K$ lies on $AC$, $N$ lies on $BC$.
Prove that $C, K, L, D, N$ lie on same circle and prove that $|KN|=|CL|$
I think that I need to do something with quadrilateral $CKDN$. I got that
$$|KN|=sqrt{|CK|^2+|CN|^2}=sqrt{|DK|^2+|DN|^2}$$
$$angle CKD+angle CND=180°$$
I also tried to express sides with the angle bisector theorem, but I don't know how to continue / what I need to solve this. How can I solve this problem?
geometry euclidean-geometry triangle circle problem-solving
asked Dec 2 '18 at 22:14
PeroPero
1207
$endgroup$
add a comment |
$begingroup$
Right triangle $Delta ABC$ ($angle ACB=90°$). The following is constructed: from point $C$ altitude $CD$, angle bisector $CL$ of $angle ACB$, angle bisector $DK$ of $angle ADC$, angle bisector $DN$ of $angle BDC$.
$D, L$ lie on $AB$, $K$ lies on $AC$, $N$ lies on $BC$.
Prove that $C, K, L, D, N$ lie on same circle and prove that $|KN|=|CL|$
I think that I need to do something with quadrilateral $CKDN$. I got that
$$|KN|=sqrt{|CK|^2+|CN|^2}=sqrt{|DK|^2+|DN|^2}$$
$$angle CKD+angle CND=180°$$
I also tried to express sides with the angle bisector theorem, but I don't know how to continue / what I need to solve this. How can I solve this problem?
geometry euclidean-geometry triangle circle problem-solving
asked Dec 2 '18 at 22:14
PeroPero
1207
$endgroup$
$begingroup$
No they aren't.
$endgroup$
– Pero
Dec 3 '18 at 12:54
add a comment |
$begingroup$
Right triangle $Delta ABC$ ($angle ACB=90°$). The following is constructed: from point $C$ altitude $CD$, angle bisector $CL$ of $angle ACB$, angle bisector $DK$ of $angle ADC$, angle bisector $DN$ of $angle BDC$.
$D, L$ lie on $AB$, $K$ lies on $AC$, $N$ lies on $BC$.
Prove that $C, K, L, D, N$ lie on same circle and prove that $|KN|=|CL|$
I think that I need to do something with quadrilateral $CKDN$. I got that
$$|KN|=sqrt{|CK|^2+|CN|^2}=sqrt{|DK|^2+|DN|^2}$$
$$angle CKD+angle CND=180°$$
I also tried to express sides with the angle bisector theorem, but I don't know how to continue / what I need to solve this. How can I solve this problem?
geometry euclidean-geometry triangle circle problem-solving
asked Dec 2 '18 at 22:14
PeroPero
1207
$endgroup$
Right triangle $Delta ABC$ ($angle ACB=90°$). The following is constructed: from point $C$ altitude $CD$, angle bisector $CL$ of $angle ACB$, angle bisector $DK$ of $angle ADC$, angle bisector $DN$ of $angle BDC$.
$D, L$ lie on $AB$, $K$ lies on $AC$, $N$ lies on $BC$.
Prove that $C, K, L, D, N$ lie on same circle and prove that $|KN|=|CL|$
I think that I need to do something with quadrilateral $CKDN$. I got that
$$|KN|=sqrt{|CK|^2+|CN|^2}=sqrt{|DK|^2+|DN|^2}$$
$$angle CKD+angle CND=180°$$
I also tried to express sides with the angle bisector theorem, but I don't know how to continue / what I need to solve this. How can I solve this problem?
geometry euclidean-geometry triangle circle problem-solving
geometry euclidean-geometry triangle circle problem-solving
asked Dec 2 '18 at 22:14
PeroPero
1207
asked Dec 2 '18 at 22:14
PeroPero
1207
edited Dec 3 '18 at 14:05
user593746
asked Dec 2 '18 at 22:14
PeroPero
1207
asked Dec 2 '18 at 22:14
PeroPero
1207
asked Dec 2 '18 at 22:14
PeroPero
1207
1207
$begingroup$
No they aren't.
$endgroup$
– Pero
Dec 3 '18 at 12:54
add a comment |
$begingroup$
No they aren't.
$endgroup$
– Pero
Dec 3 '18 at 12:54
$begingroup$
No they aren't.
$endgroup$
– Pero
Dec 3 '18 at 12:54
$begingroup$
No they aren't.
$endgroup$
– Pero
Dec 3 '18 at 12:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, since $CD perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles
$angle , ADC$ and $angle , BDC$, then $$angle , KDN = angle , KDC + angle , NDC = 45^{circ} + 45^{circ} = 90^{circ}$$
However, $angle , KCN = 90^{circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL, || , CB$ and $NL, || , CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $Delta , ADC$, we apply the theorem that
$$frac{AK}{KC} = frac{AD}{DC}$$ But triangles $Delta , ACD$ is similar to $Delta , ABC$ so $$frac{AD}{DC} = frac{AC}{CB}$$ so
$$frac{AK}{KC} = frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $Delta, ABC$
we have that
$$frac{AC}{CB} = frac{AL}{LB}$$ so consequently
$$frac{AK}{KC} = frac{AL}{LB}$$
which by Thales' intercept theorem implies that $KL , || , CB$. Analogously, one can show that $NL , || , CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
$endgroup$
$begingroup$
What is $angle ALLB$? (why not just $angle ALB$?)
$endgroup$
– Pero
Dec 3 '18 at 23:16
$begingroup$
@Pero This is a typo. It is not an angle, it was suppose to be a fraction.
$endgroup$
– Futurologist
Dec 3 '18 at 23:57
$begingroup$
Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$)
$endgroup$
– Pero
Dec 4 '18 at 0:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, since $CD perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles
$angle , ADC$ and $angle , BDC$, then $$angle , KDN = angle , KDC + angle , NDC = 45^{circ} + 45^{circ} = 90^{circ}$$
However, $angle , KCN = 90^{circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL, || , CB$ and $NL, || , CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $Delta , ADC$, we apply the theorem that
$$frac{AK}{KC} = frac{AD}{DC}$$ But triangles $Delta , ACD$ is similar to $Delta , ABC$ so $$frac{AD}{DC} = frac{AC}{CB}$$ so
$$frac{AK}{KC} = frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $Delta, ABC$
we have that
$$frac{AC}{CB} = frac{AL}{LB}$$ so consequently
$$frac{AK}{KC} = frac{AL}{LB}$$
which by Thales' intercept theorem implies that $KL , || , CB$. Analogously, one can show that $NL , || , CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
$endgroup$
$begingroup$
What is $angle ALLB$? (why not just $angle ALB$?)
$endgroup$
– Pero
Dec 3 '18 at 23:16
$begingroup$
@Pero This is a typo. It is not an angle, it was suppose to be a fraction.
$endgroup$
– Futurologist
Dec 3 '18 at 23:57
$begingroup$
Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$)
$endgroup$
– Pero
Dec 4 '18 at 0:04
add a comment |
$begingroup$
First, since $CD perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles
$angle , ADC$ and $angle , BDC$, then $$angle , KDN = angle , KDC + angle , NDC = 45^{circ} + 45^{circ} = 90^{circ}$$
However, $angle , KCN = 90^{circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL, || , CB$ and $NL, || , CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $Delta , ADC$, we apply the theorem that
$$frac{AK}{KC} = frac{AD}{DC}$$ But triangles $Delta , ACD$ is similar to $Delta , ABC$ so $$frac{AD}{DC} = frac{AC}{CB}$$ so
$$frac{AK}{KC} = frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $Delta, ABC$
we have that
$$frac{AC}{CB} = frac{AL}{LB}$$ so consequently
$$frac{AK}{KC} = frac{AL}{LB}$$
which by Thales' intercept theorem implies that $KL , || , CB$. Analogously, one can show that $NL , || , CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
$endgroup$
$begingroup$
What is $angle ALLB$? (why not just $angle ALB$?)
$endgroup$
– Pero
Dec 3 '18 at 23:16
$begingroup$
@Pero This is a typo. It is not an angle, it was suppose to be a fraction.
$endgroup$
– Futurologist
Dec 3 '18 at 23:57
$begingroup$
Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$)
$endgroup$
– Pero
Dec 4 '18 at 0:04
add a comment |
$begingroup$
First, since $CD perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles
$angle , ADC$ and $angle , BDC$, then $$angle , KDN = angle , KDC + angle , NDC = 45^{circ} + 45^{circ} = 90^{circ}$$
However, $angle , KCN = 90^{circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL, || , CB$ and $NL, || , CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $Delta , ADC$, we apply the theorem that
$$frac{AK}{KC} = frac{AD}{DC}$$ But triangles $Delta , ACD$ is similar to $Delta , ABC$ so $$frac{AD}{DC} = frac{AC}{CB}$$ so
$$frac{AK}{KC} = frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $Delta, ABC$
we have that
$$frac{AC}{CB} = frac{AL}{LB}$$ so consequently
$$frac{AK}{KC} = frac{AL}{LB}$$
which by Thales' intercept theorem implies that $KL , || , CB$. Analogously, one can show that $NL , || , CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
$endgroup$
First, since $CD perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles
$angle , ADC$ and $angle , BDC$, then $$angle , KDN = angle , KDC + angle , NDC = 45^{circ} + 45^{circ} = 90^{circ}$$
However, $angle , KCN = 90^{circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL, || , CB$ and $NL, || , CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $Delta , ADC$, we apply the theorem that
$$frac{AK}{KC} = frac{AD}{DC}$$ But triangles $Delta , ACD$ is similar to $Delta , ABC$ so $$frac{AD}{DC} = frac{AC}{CB}$$ so
$$frac{AK}{KC} = frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $Delta, ABC$
we have that
$$frac{AC}{CB} = frac{AL}{LB}$$ so consequently
$$frac{AK}{KC} = frac{AL}{LB}$$
which by Thales' intercept theorem implies that $KL , || , CB$. Analogously, one can show that $NL , || , CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
edited Dec 4 '18 at 5:22
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
answered Dec 3 '18 at 14:33
FuturologistFuturologist
7,3062520
7,3062520
$begingroup$
What is $angle ALLB$? (why not just $angle ALB$?)
$endgroup$
– Pero
Dec 3 '18 at 23:16
$begingroup$
@Pero This is a typo. It is not an angle, it was suppose to be a fraction.
$endgroup$
– Futurologist
Dec 3 '18 at 23:57
$begingroup$
Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$)
$endgroup$
– Pero
Dec 4 '18 at 0:04
add a comment |
$begingroup$
What is $angle ALLB$? (why not just $angle ALB$?)
$endgroup$
– Pero
Dec 3 '18 at 23:16
$begingroup$
@Pero This is a typo. It is not an angle, it was suppose to be a fraction.
$endgroup$
– Futurologist
Dec 3 '18 at 23:57
$begingroup$
Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$)
$endgroup$
– Pero
Dec 4 '18 at 0:04
$begingroup$
What is $angle ALLB$? (why not just $angle ALB$?)
$endgroup$
– Pero
Dec 3 '18 at 23:16
$begingroup$
What is $angle ALLB$? (why not just $angle ALB$?)
$endgroup$
– Pero
Dec 3 '18 at 23:16
$begingroup$
@Pero This is a typo. It is not an angle, it was suppose to be a fraction.
$endgroup$
– Futurologist
Dec 3 '18 at 23:57
$begingroup$
@Pero This is a typo. It is not an angle, it was suppose to be a fraction.
$endgroup$
– Futurologist
Dec 3 '18 at 23:57
$begingroup$
Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$)
$endgroup$
– Pero
Dec 4 '18 at 0:04
$begingroup$
Oh I see. I think you also have a typo at the beginning (3rd row). I think it should be $CKDN$ is inscribed in circle (not $CKDL$)
$endgroup$
– Pero
Dec 4 '18 at 0:04
add a comment |
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$begingroup$
No they aren't.
$endgroup$
– Pero
Dec 3 '18 at 12:54