Kiselev's Geometry Problem 51
$begingroup$
I am going through Kiselev's Geometry, Book I.
I am having trouble with problem 51.
The problem states:
Give an example that disproves the proposition: "If the bisectors of two angles with a common vertex are perpendicular, then the angles are supplementary." Is the converse proposition true?
His definition of a supplementary angle goes as follows:
Two angles [...] are called supplementary if they have one common side, and their remaining two sides form continuations of each other. [...] the sum of two supplementary angles is 180°.
But I cannot see how to disprove that proposition using his definition. Is it something lost in translation or is it something that I am missing?
geometry euclidean-geometry
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
$endgroup$
add a comment |
$begingroup$
I am going through Kiselev's Geometry, Book I.
I am having trouble with problem 51.
The problem states:
Give an example that disproves the proposition: "If the bisectors of two angles with a common vertex are perpendicular, then the angles are supplementary." Is the converse proposition true?
His definition of a supplementary angle goes as follows:
Two angles [...] are called supplementary if they have one common side, and their remaining two sides form continuations of each other. [...] the sum of two supplementary angles is 180°.
But I cannot see how to disprove that proposition using his definition. Is it something lost in translation or is it something that I am missing?
geometry euclidean-geometry
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
$endgroup$
add a comment |
$begingroup$
I am going through Kiselev's Geometry, Book I.
I am having trouble with problem 51.
The problem states:
Give an example that disproves the proposition: "If the bisectors of two angles with a common vertex are perpendicular, then the angles are supplementary." Is the converse proposition true?
His definition of a supplementary angle goes as follows:
Two angles [...] are called supplementary if they have one common side, and their remaining two sides form continuations of each other. [...] the sum of two supplementary angles is 180°.
But I cannot see how to disprove that proposition using his definition. Is it something lost in translation or is it something that I am missing?
geometry euclidean-geometry
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
$endgroup$
I am going through Kiselev's Geometry, Book I.
I am having trouble with problem 51.
The problem states:
Give an example that disproves the proposition: "If the bisectors of two angles with a common vertex are perpendicular, then the angles are supplementary." Is the converse proposition true?
His definition of a supplementary angle goes as follows:
Two angles [...] are called supplementary if they have one common side, and their remaining two sides form continuations of each other. [...] the sum of two supplementary angles is 180°.
But I cannot see how to disprove that proposition using his definition. Is it something lost in translation or is it something that I am missing?
geometry euclidean-geometry
geometry euclidean-geometry
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
asked Nov 28 '18 at 18:58
Júlio CezarJúlio Cezar
102
102
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Does this picture provide some help?
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
$endgroup$
$begingroup$
Yes, a lot, actually. I take that the converse proposition would then be "If two angles with a common vertex are supplementary, then the bisectors are perpendicular." and would be true?
$endgroup$
– Júlio Cezar
Nov 28 '18 at 19:19
add a comment |
$begingroup$
Hint: Angles can share a vertex without sharing a side.
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017549%2fkiselevs-geometry-problem-51%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Does this picture provide some help?
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
$endgroup$
$begingroup$
Yes, a lot, actually. I take that the converse proposition would then be "If two angles with a common vertex are supplementary, then the bisectors are perpendicular." and would be true?
$endgroup$
– Júlio Cezar
Nov 28 '18 at 19:19
add a comment |
$begingroup$
Does this picture provide some help?
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
$endgroup$
$begingroup$
Yes, a lot, actually. I take that the converse proposition would then be "If two angles with a common vertex are supplementary, then the bisectors are perpendicular." and would be true?
$endgroup$
– Júlio Cezar
Nov 28 '18 at 19:19
add a comment |
$begingroup$
Does this picture provide some help?
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
$endgroup$
Does this picture provide some help?
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
answered Nov 28 '18 at 19:06
Vasily MitchVasily Mitch
1,35837
1,35837
$begingroup$
Yes, a lot, actually. I take that the converse proposition would then be "If two angles with a common vertex are supplementary, then the bisectors are perpendicular." and would be true?
$endgroup$
– Júlio Cezar
Nov 28 '18 at 19:19
add a comment |
$begingroup$
Yes, a lot, actually. I take that the converse proposition would then be "If two angles with a common vertex are supplementary, then the bisectors are perpendicular." and would be true?
$endgroup$
– Júlio Cezar
Nov 28 '18 at 19:19
$begingroup$
Yes, a lot, actually. I take that the converse proposition would then be "If two angles with a common vertex are supplementary, then the bisectors are perpendicular." and would be true?
$endgroup$
– Júlio Cezar
Nov 28 '18 at 19:19
$begingroup$
Yes, a lot, actually. I take that the converse proposition would then be "If two angles with a common vertex are supplementary, then the bisectors are perpendicular." and would be true?
$endgroup$
– Júlio Cezar
Nov 28 '18 at 19:19
add a comment |
$begingroup$
Hint: Angles can share a vertex without sharing a side.
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
$endgroup$
add a comment |
$begingroup$
Hint: Angles can share a vertex without sharing a side.
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
$endgroup$
add a comment |
$begingroup$
Hint: Angles can share a vertex without sharing a side.
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
$endgroup$
Hint: Angles can share a vertex without sharing a side.
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
answered Nov 28 '18 at 19:05
ArthurArthur
112k7107191
112k7107191
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017549%2fkiselevs-geometry-problem-51%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
