Mixing
What is the value of AD+CD where ABC is an isosceles triangle, D bisects angle ACB, BC = 2017 unit?
$begingroup$
$ABC$ is an isosceles ($AB = AC$) and $angle A = 100^{circ}$. An point $D$ is on $AB$ so that $CD$ angle bisector $angle ACB$. If $BC= 2017$ then calculate $AD+CD$.
Source: Bangladesh Math Olympiad 2017 junior category.
geometry contest-math euclidean-geometry
edited Jan 9 at 20:28
greedoid
40.6k1149100
asked Jan 9 at 14:00
ShromiShromi
1199
$endgroup$
add a comment |
$begingroup$
$ABC$ is an isosceles ($AB = AC$) and $angle A = 100^{circ}$. An point $D$ is on $AB$ so that $CD$ angle bisector $angle ACB$. If $BC= 2017$ then calculate $AD+CD$.
Source: Bangladesh Math Olympiad 2017 junior category.
geometry contest-math euclidean-geometry
edited Jan 9 at 20:28
greedoid
40.6k1149100
asked Jan 9 at 14:00
ShromiShromi
1199
$endgroup$
1
$begingroup$
all angles are known: do you know sines law ?
$endgroup$
– G Cab
Jan 9 at 16:45
$begingroup$
Thanks I've just learned that. But I am confused that how am I supposed to figure it out without calculator. (Sine 40 degree) Could you please help me?
$endgroup$
– Shromi
Jan 9 at 17:00
1
$begingroup$
didn't get through the solution, but probably you are supposed to use the fact that $40 = 100-60$
$endgroup$
– G Cab
Jan 9 at 17:05
add a comment |
$begingroup$
$ABC$ is an isosceles ($AB = AC$) and $angle A = 100^{circ}$. An point $D$ is on $AB$ so that $CD$ angle bisector $angle ACB$. If $BC= 2017$ then calculate $AD+CD$.
Source: Bangladesh Math Olympiad 2017 junior category.
geometry contest-math euclidean-geometry
edited Jan 9 at 20:28
greedoid
40.6k1149100
asked Jan 9 at 14:00
ShromiShromi
1199
$endgroup$
$ABC$ is an isosceles ($AB = AC$) and $angle A = 100^{circ}$. An point $D$ is on $AB$ so that $CD$ angle bisector $angle ACB$. If $BC= 2017$ then calculate $AD+CD$.
Source: Bangladesh Math Olympiad 2017 junior category.
geometry contest-math euclidean-geometry
geometry contest-math euclidean-geometry
edited Jan 9 at 20:28
greedoid
40.6k1149100
asked Jan 9 at 14:00
ShromiShromi
1199
edited Jan 9 at 20:28
greedoid
40.6k1149100
asked Jan 9 at 14:00
ShromiShromi
1199
edited Jan 9 at 20:28
greedoid
40.6k1149100
edited Jan 9 at 20:28
greedoid
40.6k1149100
edited Jan 9 at 20:28
greedoid
40.6k1149100
40.6k1149100
asked Jan 9 at 14:00
ShromiShromi
1199
asked Jan 9 at 14:00
ShromiShromi
1199
asked Jan 9 at 14:00
ShromiShromi
1199
1199
1
$begingroup$
all angles are known: do you know sines law ?
$endgroup$
– G Cab
Jan 9 at 16:45
$begingroup$
Thanks I've just learned that. But I am confused that how am I supposed to figure it out without calculator. (Sine 40 degree) Could you please help me?
$endgroup$
– Shromi
Jan 9 at 17:00
1
$begingroup$
didn't get through the solution, but probably you are supposed to use the fact that $40 = 100-60$
$endgroup$
– G Cab
Jan 9 at 17:05
add a comment |
1
$begingroup$
all angles are known: do you know sines law ?
$endgroup$
– G Cab
Jan 9 at 16:45
$begingroup$
Thanks I've just learned that. But I am confused that how am I supposed to figure it out without calculator. (Sine 40 degree) Could you please help me?
$endgroup$
– Shromi
Jan 9 at 17:00
1
$begingroup$
didn't get through the solution, but probably you are supposed to use the fact that $40 = 100-60$
$endgroup$
– G Cab
Jan 9 at 17:05
1
1
$begingroup$
all angles are known: do you know sines law ?
$endgroup$
– G Cab
Jan 9 at 16:45
$begingroup$
all angles are known: do you know sines law ?
$endgroup$
– G Cab
Jan 9 at 16:45
$begingroup$
Thanks I've just learned that. But I am confused that how am I supposed to figure it out without calculator. (Sine 40 degree) Could you please help me?
$endgroup$
– Shromi
Jan 9 at 17:00
$begingroup$
Thanks I've just learned that. But I am confused that how am I supposed to figure it out without calculator. (Sine 40 degree) Could you please help me?
$endgroup$
– Shromi
Jan 9 at 17:00
1
1
$begingroup$
didn't get through the solution, but probably you are supposed to use the fact that $40 = 100-60$
$endgroup$
– G Cab
Jan 9 at 17:05
$begingroup$
didn't get through the solution, but probably you are supposed to use the fact that $40 = 100-60$
$endgroup$
– G Cab
Jan 9 at 17:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $E$ be on $BC$ so $CE = CD$. Then $angle DEC = 80^{circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$acdot (a-d) = bcdot (b-x);;;;;;;;;;;;(1)$$
By angle bisector theorem we have $${b-xover x} = {aover b}implies b(b-x)= ax;;;;;;;;(2)$$
By (1) and (2) we have $$a(a-d)= ax implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let $E$ be on $BC$ so $CE = CD$. Then $angle DEC = 80^{circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$acdot (a-d) = bcdot (b-x);;;;;;;;;;;;(1)$$
By angle bisector theorem we have $${b-xover x} = {aover b}implies b(b-x)= ax;;;;;;;;(2)$$
By (1) and (2) we have $$a(a-d)= ax implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
$endgroup$
add a comment |
$begingroup$
Let $E$ be on $BC$ so $CE = CD$. Then $angle DEC = 80^{circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$acdot (a-d) = bcdot (b-x);;;;;;;;;;;;(1)$$
By angle bisector theorem we have $${b-xover x} = {aover b}implies b(b-x)= ax;;;;;;;;(2)$$
By (1) and (2) we have $$a(a-d)= ax implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
$endgroup$
add a comment |
$begingroup$
Let $E$ be on $BC$ so $CE = CD$. Then $angle DEC = 80^{circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$acdot (a-d) = bcdot (b-x);;;;;;;;;;;;(1)$$
By angle bisector theorem we have $${b-xover x} = {aover b}implies b(b-x)= ax;;;;;;;;(2)$$
By (1) and (2) we have $$a(a-d)= ax implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
$endgroup$
Let $E$ be on $BC$ so $CE = CD$. Then $angle DEC = 80^{circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$acdot (a-d) = bcdot (b-x);;;;;;;;;;;;(1)$$
By angle bisector theorem we have $${b-xover x} = {aover b}implies b(b-x)= ax;;;;;;;;(2)$$
By (1) and (2) we have $$a(a-d)= ax implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
answered Jan 9 at 18:39
greedoidgreedoid
40.6k1149100
40.6k1149100
add a comment |
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1
$begingroup$
all angles are known: do you know sines law ?
$endgroup$
– G Cab
Jan 9 at 16:45
$begingroup$
Thanks I've just learned that. But I am confused that how am I supposed to figure it out without calculator. (Sine 40 degree) Could you please help me?
$endgroup$
– Shromi
Jan 9 at 17:00
1
$begingroup$
didn't get through the solution, but probably you are supposed to use the fact that $40 = 100-60$
$endgroup$
– G Cab
Jan 9 at 17:05