Mixing
Prove that in the given pentagon ABCDE AB||CE
$begingroup$
In a convex pentagon $ABCDE$ we have $BCparallel AD$, $CDparallel BE$, $DEparallel AC$ and $AEparallel BD$.
Prove that $ABparallel CE$.
geometry
asked Jan 16 at 21:10
DoodDood
487
$endgroup$
add a comment |
$begingroup$
In a convex pentagon $ABCDE$ we have $BCparallel AD$, $CDparallel BE$, $DEparallel AC$ and $AEparallel BD$.
Prove that $ABparallel CE$.
geometry
asked Jan 16 at 21:10
DoodDood
487
$endgroup$
$begingroup$
Please make the edit. AE || BD must be included
$endgroup$
– Rahuboy
Jan 17 at 10:03
add a comment |
$begingroup$
In a convex pentagon $ABCDE$ we have $BCparallel AD$, $CDparallel BE$, $DEparallel AC$ and $AEparallel BD$.
Prove that $ABparallel CE$.
geometry
asked Jan 16 at 21:10
DoodDood
487
$endgroup$
In a convex pentagon $ABCDE$ we have $BCparallel AD$, $CDparallel BE$, $DEparallel AC$ and $AEparallel BD$.
Prove that $ABparallel CE$.
geometry
geometry
asked Jan 16 at 21:10
DoodDood
487
asked Jan 16 at 21:10
DoodDood
487
edited Jan 17 at 10:06
Dood
asked Jan 16 at 21:10
DoodDood
487
asked Jan 16 at 21:10
DoodDood
487
asked Jan 16 at 21:10
DoodDood
487
487
$begingroup$
Please make the edit. AE || BD must be included
$endgroup$
– Rahuboy
Jan 17 at 10:03
add a comment |
$begingroup$
Please make the edit. AE || BD must be included
$endgroup$
– Rahuboy
Jan 17 at 10:03
$begingroup$
Please make the edit. AE || BD must be included
$endgroup$
– Rahuboy
Jan 17 at 10:03
$begingroup$
Please make the edit. AE || BD must be included
$endgroup$
– Rahuboy
Jan 17 at 10:03
add a comment |
1 Answer
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$begingroup$
Consider the above figure. All the angles of the same color are equal , using properties of parallel lines. Also , using the fact that the opposite sides of parellelograms are equal , all the line segments of the same color are equal. We are required to prove that $bf{ABparallel CE}$
First , we join $M_1J_1$ , $N_1J_1$ , $N_1K_1$ and $M_1L_1$. Using the converse of Thales’ Theorem , we can prove that they are parallel to segments $EA&N_1L_1$,$BC & M_1K_1$, $AC$ and $BE$ respectively. Denote the blue segments by $b$ , the yellow segments by $y$ , $J_1L_1$ by $x$ and $K_1J_1$ by $z$ . Applying Thales’ Theorem to the required $triangle$s , we have $$frac{EM_1}{M_1C}=frac{b}{x+b}=frac{x}{b}=frac{M_1N_1}{N_1C}$$ $$frac{CN_1}{N_1E}=frac{y}{z+y}=frac{z}{y}=frac{N_1M_1}{EM_1}$$
From these two equations , we get $EM_1=N_1C$ And from this , we get $$frac{x}{b}=frac{z}{y}$$ Therefore , $$frac{J_1E}{J_1B}=frac{J_1C}{J_1A}$$ Therefore , $triangle J_1AB sim triangle J_1CE $ by $SAS$ similarity.
It follows that $bf{CE parallel AB}$
answered Jan 17 at 9:59
RahuboyRahuboy
64511
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider the above figure. All the angles of the same color are equal , using properties of parallel lines. Also , using the fact that the opposite sides of parellelograms are equal , all the line segments of the same color are equal. We are required to prove that $bf{ABparallel CE}$
First , we join $M_1J_1$ , $N_1J_1$ , $N_1K_1$ and $M_1L_1$. Using the converse of Thales’ Theorem , we can prove that they are parallel to segments $EA&N_1L_1$,$BC & M_1K_1$, $AC$ and $BE$ respectively. Denote the blue segments by $b$ , the yellow segments by $y$ , $J_1L_1$ by $x$ and $K_1J_1$ by $z$ . Applying Thales’ Theorem to the required $triangle$s , we have $$frac{EM_1}{M_1C}=frac{b}{x+b}=frac{x}{b}=frac{M_1N_1}{N_1C}$$ $$frac{CN_1}{N_1E}=frac{y}{z+y}=frac{z}{y}=frac{N_1M_1}{EM_1}$$
From these two equations , we get $EM_1=N_1C$ And from this , we get $$frac{x}{b}=frac{z}{y}$$ Therefore , $$frac{J_1E}{J_1B}=frac{J_1C}{J_1A}$$ Therefore , $triangle J_1AB sim triangle J_1CE $ by $SAS$ similarity.
It follows that $bf{CE parallel AB}$
answered Jan 17 at 9:59
RahuboyRahuboy
64511
$endgroup$
add a comment |
$begingroup$
Consider the above figure. All the angles of the same color are equal , using properties of parallel lines. Also , using the fact that the opposite sides of parellelograms are equal , all the line segments of the same color are equal. We are required to prove that $bf{ABparallel CE}$
First , we join $M_1J_1$ , $N_1J_1$ , $N_1K_1$ and $M_1L_1$. Using the converse of Thales’ Theorem , we can prove that they are parallel to segments $EA&N_1L_1$,$BC & M_1K_1$, $AC$ and $BE$ respectively. Denote the blue segments by $b$ , the yellow segments by $y$ , $J_1L_1$ by $x$ and $K_1J_1$ by $z$ . Applying Thales’ Theorem to the required $triangle$s , we have $$frac{EM_1}{M_1C}=frac{b}{x+b}=frac{x}{b}=frac{M_1N_1}{N_1C}$$ $$frac{CN_1}{N_1E}=frac{y}{z+y}=frac{z}{y}=frac{N_1M_1}{EM_1}$$
From these two equations , we get $EM_1=N_1C$ And from this , we get $$frac{x}{b}=frac{z}{y}$$ Therefore , $$frac{J_1E}{J_1B}=frac{J_1C}{J_1A}$$ Therefore , $triangle J_1AB sim triangle J_1CE $ by $SAS$ similarity.
It follows that $bf{CE parallel AB}$
answered Jan 17 at 9:59
RahuboyRahuboy
64511
$endgroup$
add a comment |
$begingroup$
Consider the above figure. All the angles of the same color are equal , using properties of parallel lines. Also , using the fact that the opposite sides of parellelograms are equal , all the line segments of the same color are equal. We are required to prove that $bf{ABparallel CE}$
First , we join $M_1J_1$ , $N_1J_1$ , $N_1K_1$ and $M_1L_1$. Using the converse of Thales’ Theorem , we can prove that they are parallel to segments $EA&N_1L_1$,$BC & M_1K_1$, $AC$ and $BE$ respectively. Denote the blue segments by $b$ , the yellow segments by $y$ , $J_1L_1$ by $x$ and $K_1J_1$ by $z$ . Applying Thales’ Theorem to the required $triangle$s , we have $$frac{EM_1}{M_1C}=frac{b}{x+b}=frac{x}{b}=frac{M_1N_1}{N_1C}$$ $$frac{CN_1}{N_1E}=frac{y}{z+y}=frac{z}{y}=frac{N_1M_1}{EM_1}$$
From these two equations , we get $EM_1=N_1C$ And from this , we get $$frac{x}{b}=frac{z}{y}$$ Therefore , $$frac{J_1E}{J_1B}=frac{J_1C}{J_1A}$$ Therefore , $triangle J_1AB sim triangle J_1CE $ by $SAS$ similarity.
It follows that $bf{CE parallel AB}$
answered Jan 17 at 9:59
RahuboyRahuboy
64511
$endgroup$
Consider the above figure. All the angles of the same color are equal , using properties of parallel lines. Also , using the fact that the opposite sides of parellelograms are equal , all the line segments of the same color are equal. We are required to prove that $bf{ABparallel CE}$
First , we join $M_1J_1$ , $N_1J_1$ , $N_1K_1$ and $M_1L_1$. Using the converse of Thales’ Theorem , we can prove that they are parallel to segments $EA&N_1L_1$,$BC & M_1K_1$, $AC$ and $BE$ respectively. Denote the blue segments by $b$ , the yellow segments by $y$ , $J_1L_1$ by $x$ and $K_1J_1$ by $z$ . Applying Thales’ Theorem to the required $triangle$s , we have $$frac{EM_1}{M_1C}=frac{b}{x+b}=frac{x}{b}=frac{M_1N_1}{N_1C}$$ $$frac{CN_1}{N_1E}=frac{y}{z+y}=frac{z}{y}=frac{N_1M_1}{EM_1}$$
From these two equations , we get $EM_1=N_1C$ And from this , we get $$frac{x}{b}=frac{z}{y}$$ Therefore , $$frac{J_1E}{J_1B}=frac{J_1C}{J_1A}$$ Therefore , $triangle J_1AB sim triangle J_1CE $ by $SAS$ similarity.
It follows that $bf{CE parallel AB}$
answered Jan 17 at 9:59
RahuboyRahuboy
64511
edited Jan 17 at 10:08
answered Jan 17 at 9:59
RahuboyRahuboy
64511
answered Jan 17 at 9:59
RahuboyRahuboy
64511
answered Jan 17 at 9:59
RahuboyRahuboy
64511
64511
add a comment |
add a comment |
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$begingroup$
Please make the edit. AE || BD must be included
$endgroup$
– Rahuboy
Jan 17 at 10:03