Mixing
Prove that: BD/DC = AR/AS
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In a triangle ABC, internal bisector of angle A intersects side BC at D. R and S are circumcentres of triangle ABD and triangle ADC respectively. Then prove that, BD/DC = AR/AS.
euclidean-geometry
asked Jan 15 at 10:32
RB MCPERB MCPE
262
$endgroup$
add a comment |
$begingroup$
In a triangle ABC, internal bisector of angle A intersects side BC at D. R and S are circumcentres of triangle ABD and triangle ADC respectively. Then prove that, BD/DC = AR/AS.
euclidean-geometry
asked Jan 15 at 10:32
RB MCPERB MCPE
262
$endgroup$
1
$begingroup$
Welcome to MSE. What have you tried so far? Are you stuck on some point?
$endgroup$
– Andrei
Jan 15 at 10:39
$begingroup$
I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it
$endgroup$
– RB MCPE
Jan 15 at 10:56
add a comment |
$begingroup$
In a triangle ABC, internal bisector of angle A intersects side BC at D. R and S are circumcentres of triangle ABD and triangle ADC respectively. Then prove that, BD/DC = AR/AS.
euclidean-geometry
asked Jan 15 at 10:32
RB MCPERB MCPE
262
$endgroup$
In a triangle ABC, internal bisector of angle A intersects side BC at D. R and S are circumcentres of triangle ABD and triangle ADC respectively. Then prove that, BD/DC = AR/AS.
euclidean-geometry
euclidean-geometry
asked Jan 15 at 10:32
RB MCPERB MCPE
262
asked Jan 15 at 10:32
RB MCPERB MCPE
262
asked Jan 15 at 10:32
RB MCPERB MCPE
262
asked Jan 15 at 10:32
RB MCPERB MCPE
262
asked Jan 15 at 10:32
RB MCPERB MCPE
262
262
1
$begingroup$
Welcome to MSE. What have you tried so far? Are you stuck on some point?
$endgroup$
– Andrei
Jan 15 at 10:39
$begingroup$
I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it
$endgroup$
– RB MCPE
Jan 15 at 10:56
add a comment |
1
$begingroup$
Welcome to MSE. What have you tried so far? Are you stuck on some point?
$endgroup$
– Andrei
Jan 15 at 10:39
$begingroup$
I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it
$endgroup$
– RB MCPE
Jan 15 at 10:56
1
1
$begingroup$
Welcome to MSE. What have you tried so far? Are you stuck on some point?
$endgroup$
– Andrei
Jan 15 at 10:39
$begingroup$
Welcome to MSE. What have you tried so far? Are you stuck on some point?
$endgroup$
– Andrei
Jan 15 at 10:39
$begingroup$
I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it
$endgroup$
– RB MCPE
Jan 15 at 10:56
$begingroup$
I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it
$endgroup$
– RB MCPE
Jan 15 at 10:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Firstly by angle chasing we can show that the 2 isosceles triangles $triangle ARB equiv triangle ASC$. Now it is well known that $frac{BD}{DC}=frac{AB}{AC} (1)$. But since $triangle ARB equiv triangle ASC$ then $frac{AB}{AC}=frac{AR}{AS} (2)$. Now combining $(1)$ and $(2)$ gives us the desired result. $square$
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
$endgroup$
$begingroup$
Can u help me in proving $triangle ABC$ similar to $triangle ARS$ ?
$endgroup$
– RB MCPE
Jan 15 at 11:27
$begingroup$
Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $triangle ASR$ and $triangle ABC$ it follows that they are similar.
$endgroup$
– Sota Antonino
Jan 15 at 12:36
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Firstly by angle chasing we can show that the 2 isosceles triangles $triangle ARB equiv triangle ASC$. Now it is well known that $frac{BD}{DC}=frac{AB}{AC} (1)$. But since $triangle ARB equiv triangle ASC$ then $frac{AB}{AC}=frac{AR}{AS} (2)$. Now combining $(1)$ and $(2)$ gives us the desired result. $square$
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
$endgroup$
$begingroup$
Can u help me in proving $triangle ABC$ similar to $triangle ARS$ ?
$endgroup$
– RB MCPE
Jan 15 at 11:27
$begingroup$
Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $triangle ASR$ and $triangle ABC$ it follows that they are similar.
$endgroup$
– Sota Antonino
Jan 15 at 12:36
add a comment |
$begingroup$
Firstly by angle chasing we can show that the 2 isosceles triangles $triangle ARB equiv triangle ASC$. Now it is well known that $frac{BD}{DC}=frac{AB}{AC} (1)$. But since $triangle ARB equiv triangle ASC$ then $frac{AB}{AC}=frac{AR}{AS} (2)$. Now combining $(1)$ and $(2)$ gives us the desired result. $square$
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
$endgroup$
$begingroup$
Can u help me in proving $triangle ABC$ similar to $triangle ARS$ ?
$endgroup$
– RB MCPE
Jan 15 at 11:27
$begingroup$
Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $triangle ASR$ and $triangle ABC$ it follows that they are similar.
$endgroup$
– Sota Antonino
Jan 15 at 12:36
add a comment |
$begingroup$
Firstly by angle chasing we can show that the 2 isosceles triangles $triangle ARB equiv triangle ASC$. Now it is well known that $frac{BD}{DC}=frac{AB}{AC} (1)$. But since $triangle ARB equiv triangle ASC$ then $frac{AB}{AC}=frac{AR}{AS} (2)$. Now combining $(1)$ and $(2)$ gives us the desired result. $square$
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
$endgroup$
Firstly by angle chasing we can show that the 2 isosceles triangles $triangle ARB equiv triangle ASC$. Now it is well known that $frac{BD}{DC}=frac{AB}{AC} (1)$. But since $triangle ARB equiv triangle ASC$ then $frac{AB}{AC}=frac{AR}{AS} (2)$. Now combining $(1)$ and $(2)$ gives us the desired result. $square$
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
answered Jan 15 at 10:45
Sota AntoninoSota Antonino
887
887
$begingroup$
Can u help me in proving $triangle ABC$ similar to $triangle ARS$ ?
$endgroup$
– RB MCPE
Jan 15 at 11:27
$begingroup$
Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $triangle ASR$ and $triangle ABC$ it follows that they are similar.
$endgroup$
– Sota Antonino
Jan 15 at 12:36
add a comment |
$begingroup$
Can u help me in proving $triangle ABC$ similar to $triangle ARS$ ?
$endgroup$
– RB MCPE
Jan 15 at 11:27
$begingroup$
Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $triangle ASR$ and $triangle ABC$ it follows that they are similar.
$endgroup$
– Sota Antonino
Jan 15 at 12:36
$begingroup$
Can u help me in proving $triangle ABC$ similar to $triangle ARS$ ?
$endgroup$
– RB MCPE
Jan 15 at 11:27
$begingroup$
Can u help me in proving $triangle ABC$ similar to $triangle ARS$ ?
$endgroup$
– RB MCPE
Jan 15 at 11:27
$begingroup$
Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $triangle ASR$ and $triangle ABC$ it follows that they are similar.
$endgroup$
– Sota Antonino
Jan 15 at 12:36
$begingroup$
Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $triangle ASR$ and $triangle ABC$ it follows that they are similar.
$endgroup$
– Sota Antonino
Jan 15 at 12:36
add a comment |
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1
$begingroup$
Welcome to MSE. What have you tried so far? Are you stuck on some point?
$endgroup$
– Andrei
Jan 15 at 10:39
$begingroup$
I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it
$endgroup$
– RB MCPE
Jan 15 at 10:56