Can't figure out this triangle geometry problem
I have the following triangle:
The following information about it are given:
- ABCD is a trapezoid (AB || DC)
- EF || DC
- Q is the intersection of AC, DB, PN, & EF
Prove that EQ = QF.
Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:
$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$
$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$
I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.
How can I solve this particular problem, and how do I tackle problems of this kind more effectively?
geometry euclidean-geometry geometric-transformation
edited 2 days ago
greedoid
38.2k114797
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
add a comment |
I have the following triangle:
The following information about it are given:
- ABCD is a trapezoid (AB || DC)
- EF || DC
- Q is the intersection of AC, DB, PN, & EF
Prove that EQ = QF.
Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:
$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$
$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$
I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.
How can I solve this particular problem, and how do I tackle problems of this kind more effectively?
geometry euclidean-geometry geometric-transformation
edited 2 days ago
greedoid
38.2k114797
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
– Matteo
Jan 2 at 20:46
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
– greedoid
Jan 2 at 20:51
Q is just a point. No information about middle points is given.
– daedsidog
Jan 2 at 20:51
add a comment |
I have the following triangle:
The following information about it are given:
- ABCD is a trapezoid (AB || DC)
- EF || DC
- Q is the intersection of AC, DB, PN, & EF
Prove that EQ = QF.
Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:
$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$
$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$
I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.
How can I solve this particular problem, and how do I tackle problems of this kind more effectively?
geometry euclidean-geometry geometric-transformation
edited 2 days ago
greedoid
38.2k114797
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
I have the following triangle:
The following information about it are given:
- ABCD is a trapezoid (AB || DC)
- EF || DC
- Q is the intersection of AC, DB, PN, & EF
Prove that EQ = QF.
Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:
$$bigtriangleup text{APM} sim bigtriangleup text{EPQ} text{ and } bigtriangleup text{PMB} sim bigtriangleup text{PQF}$$
$$begin{cases} frac{AM}{EQ} = frac{PM}{PQ} \
frac{MB}{QF} = frac{PM}{PQ} end{cases}$$
I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.
How can I solve this particular problem, and how do I tackle problems of this kind more effectively?
geometry euclidean-geometry geometric-transformation
geometry euclidean-geometry geometric-transformation
edited 2 days ago
greedoid
38.2k114797
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
edited 2 days ago
greedoid
38.2k114797
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
edited 2 days ago
greedoid
38.2k114797
edited 2 days ago
greedoid
38.2k114797
edited 2 days ago
greedoid
38.2k114797
38.2k114797
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
asked Jan 2 at 20:36
daedsidogdaedsidog
26016
26016
It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
– Matteo
Jan 2 at 20:46
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
– greedoid
Jan 2 at 20:51
Q is just a point. No information about middle points is given.
– daedsidog
Jan 2 at 20:51
add a comment |
It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
– Matteo
Jan 2 at 20:46
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
– greedoid
Jan 2 at 20:51
Q is just a point. No information about middle points is given.
– daedsidog
Jan 2 at 20:51
It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
– Matteo
Jan 2 at 20:46
It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
– Matteo
Jan 2 at 20:46
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
– greedoid
Jan 2 at 20:51
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
– greedoid
Jan 2 at 20:51
Q is just a point. No information about middle points is given.
– daedsidog
Jan 2 at 20:51
Q is just a point. No information about middle points is given.
– daedsidog
Jan 2 at 20:51
add a comment |
3 Answers
3
active
oldest
votes
First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).
$$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
– daedsidog
Jan 2 at 21:01
add a comment |
Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.
Let’s first prove that $P, M, Q, N$ are aligned.
The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.
Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.
Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
By similarity twice and by Thales for $angle DPC$ we obtain:
$$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
add a comment |
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3 Answers
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).
$$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
– daedsidog
Jan 2 at 21:01
add a comment |
First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).
$$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
– daedsidog
Jan 2 at 21:01
add a comment |
First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).
$$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
First and last equality are because of triangle similarty ($BQFsim BDC$ and $AEQ sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).
$$ {QF over CD} = {QBover DB} = {QAover AC} = {EQover CD}$$
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
answered Jan 2 at 20:56
greedoidgreedoid
38.2k114797
38.2k114797
Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
– daedsidog
Jan 2 at 21:01
add a comment |
Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
– daedsidog
Jan 2 at 21:01
Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
– daedsidog
Jan 2 at 21:01
Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this?
– daedsidog
Jan 2 at 21:01
add a comment |
Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.
Let’s first prove that $P, M, Q, N$ are aligned.
The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.
Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.
Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.
Let’s first prove that $P, M, Q, N$ are aligned.
The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.
Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.
Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.
Let’s first prove that $P, M, Q, N$ are aligned.
The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.
Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.
Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.
Let’s first prove that $P, M, Q, N$ are aligned.
The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.
Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.
Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered Jan 2 at 21:03
mathcounterexamples.netmathcounterexamples.net
25.2k21953
25.2k21953
add a comment |
add a comment |
By similarity twice and by Thales for $angle DPC$ we obtain:
$$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
add a comment |
By similarity twice and by Thales for $angle DPC$ we obtain:
$$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
add a comment |
By similarity twice and by Thales for $angle DPC$ we obtain:
$$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
By similarity twice and by Thales for $angle DPC$ we obtain:
$$frac{EQ}{AB}=frac{DE}{DA}=frac{CF}{CB}=frac{QF}{AB}.$$
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
answered Jan 2 at 22:01
Michael RozenbergMichael Rozenberg
97.3k1589188
97.3k1589188
add a comment |
add a comment |
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It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)?
– Matteo
Jan 2 at 20:46
@Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem
– greedoid
Jan 2 at 20:51
Q is just a point. No information about middle points is given.
– daedsidog
Jan 2 at 20:51