Proof With Cyclic Quadrilateral and Circle
$begingroup$
Let $ABCD$ be a cyclic quadrilateral. Let $P$ be the intersection of $overline{AD}$ and $overline{BC}$, and let $Q$ be the intersection of $overline{AB}$ and $overline{CD}$. Prove that the angle bisectors of $angle DPC$ and $angle AQD$ are perpendicular.
All I know about cyclic quadrilaterals is that opposite angles add up to $180^circ.$ How can I use that to finish my proof? Thanks in advance for answering!
geometry circle angle quadrilateral
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
$endgroup$
add a comment |
$begingroup$
Let $ABCD$ be a cyclic quadrilateral. Let $P$ be the intersection of $overline{AD}$ and $overline{BC}$, and let $Q$ be the intersection of $overline{AB}$ and $overline{CD}$. Prove that the angle bisectors of $angle DPC$ and $angle AQD$ are perpendicular.
All I know about cyclic quadrilaterals is that opposite angles add up to $180^circ.$ How can I use that to finish my proof? Thanks in advance for answering!
geometry circle angle quadrilateral
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
$endgroup$
add a comment |
$begingroup$
Let $ABCD$ be a cyclic quadrilateral. Let $P$ be the intersection of $overline{AD}$ and $overline{BC}$, and let $Q$ be the intersection of $overline{AB}$ and $overline{CD}$. Prove that the angle bisectors of $angle DPC$ and $angle AQD$ are perpendicular.
All I know about cyclic quadrilaterals is that opposite angles add up to $180^circ.$ How can I use that to finish my proof? Thanks in advance for answering!
geometry circle angle quadrilateral
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
$endgroup$
Let $ABCD$ be a cyclic quadrilateral. Let $P$ be the intersection of $overline{AD}$ and $overline{BC}$, and let $Q$ be the intersection of $overline{AB}$ and $overline{CD}$. Prove that the angle bisectors of $angle DPC$ and $angle AQD$ are perpendicular.
All I know about cyclic quadrilaterals is that opposite angles add up to $180^circ.$ How can I use that to finish my proof? Thanks in advance for answering!
geometry circle angle quadrilateral
geometry circle angle quadrilateral
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
asked Sep 14 '16 at 21:18
DreamerDreamer
680517
680517
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We need a few more letters: Let the intersection between the bisectors be at $I$, let $E, F$ be where the bisector at $P$ intersects $AB$ and $CD$ respectively, and let $G,H$ be where the bisector at $Q$ intersects $BC$ and $AD$ respectively. This gives the following figure:
(In retrospect, $E$ and $G$ were unnecessary, but hey, that's how you solve geometry puzzles: You keep adding points and lines and calculate until you find the ones you really needed.)
Let, for convinience, $angle DAB = a$ and $angle ADC = d$ (assuming the quadrilateral is named so that $A$ is between $D$ and $P$, and $B$ is between $A$ and $Q$). Now we chase angles.
$angle AQD = 180^{circ} - a-d$ and $angle DPC = a-d$ (since $angle DCP = 180^circ - a$).
Since $QH$ bisects $angle AQD$, we get $angle DHQ = 90 + frac a2 - frac d2 = $ and similarily $angle DFP = 180^circ - frac{a}{2} - frac d2$. That means that in the quadrilateral $DFIH$ we know three of the angles, and they add up to
$$
angle ADC + angle DFP + angle DHQ = d + 180^circ - frac{a}{2} - frac d2 + 90^circ + frac a{2} - frac d2 = 270^circ
$$
Thus the last angle in the quadrilateral must be $90^circ$.
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
Construction:- 1) Extend CP. 2) Draw PR, the external bisector of $angle APB$.
It should be clear that $theta = theta’$, $phi = phi’$, $omega = omega’$, and $n = n'$ [ext. angle, cyclic quad.]
The logic:- We know that the internal and the external angle bisector are perpendicular to each other. Thus, we are done if we can prove that PR // QI. The question now is changed to “will $omega = alpha?$”
By exterior angle, $omega = dfrac {m + n’}{2} = dfrac {(n + 2phi) + n'}{2} = n+phi= alpha$.
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
$endgroup$
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%2f1927069%2fproof-with-cyclic-quadrilateral-and-circle%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
$begingroup$
We need a few more letters: Let the intersection between the bisectors be at $I$, let $E, F$ be where the bisector at $P$ intersects $AB$ and $CD$ respectively, and let $G,H$ be where the bisector at $Q$ intersects $BC$ and $AD$ respectively. This gives the following figure:
(In retrospect, $E$ and $G$ were unnecessary, but hey, that's how you solve geometry puzzles: You keep adding points and lines and calculate until you find the ones you really needed.)
Let, for convinience, $angle DAB = a$ and $angle ADC = d$ (assuming the quadrilateral is named so that $A$ is between $D$ and $P$, and $B$ is between $A$ and $Q$). Now we chase angles.
$angle AQD = 180^{circ} - a-d$ and $angle DPC = a-d$ (since $angle DCP = 180^circ - a$).
Since $QH$ bisects $angle AQD$, we get $angle DHQ = 90 + frac a2 - frac d2 = $ and similarily $angle DFP = 180^circ - frac{a}{2} - frac d2$. That means that in the quadrilateral $DFIH$ we know three of the angles, and they add up to
$$
angle ADC + angle DFP + angle DHQ = d + 180^circ - frac{a}{2} - frac d2 + 90^circ + frac a{2} - frac d2 = 270^circ
$$
Thus the last angle in the quadrilateral must be $90^circ$.
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
We need a few more letters: Let the intersection between the bisectors be at $I$, let $E, F$ be where the bisector at $P$ intersects $AB$ and $CD$ respectively, and let $G,H$ be where the bisector at $Q$ intersects $BC$ and $AD$ respectively. This gives the following figure:
(In retrospect, $E$ and $G$ were unnecessary, but hey, that's how you solve geometry puzzles: You keep adding points and lines and calculate until you find the ones you really needed.)
Let, for convinience, $angle DAB = a$ and $angle ADC = d$ (assuming the quadrilateral is named so that $A$ is between $D$ and $P$, and $B$ is between $A$ and $Q$). Now we chase angles.
$angle AQD = 180^{circ} - a-d$ and $angle DPC = a-d$ (since $angle DCP = 180^circ - a$).
Since $QH$ bisects $angle AQD$, we get $angle DHQ = 90 + frac a2 - frac d2 = $ and similarily $angle DFP = 180^circ - frac{a}{2} - frac d2$. That means that in the quadrilateral $DFIH$ we know three of the angles, and they add up to
$$
angle ADC + angle DFP + angle DHQ = d + 180^circ - frac{a}{2} - frac d2 + 90^circ + frac a{2} - frac d2 = 270^circ
$$
Thus the last angle in the quadrilateral must be $90^circ$.
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
$endgroup$
add a comment |
$begingroup$
We need a few more letters: Let the intersection between the bisectors be at $I$, let $E, F$ be where the bisector at $P$ intersects $AB$ and $CD$ respectively, and let $G,H$ be where the bisector at $Q$ intersects $BC$ and $AD$ respectively. This gives the following figure:
(In retrospect, $E$ and $G$ were unnecessary, but hey, that's how you solve geometry puzzles: You keep adding points and lines and calculate until you find the ones you really needed.)
Let, for convinience, $angle DAB = a$ and $angle ADC = d$ (assuming the quadrilateral is named so that $A$ is between $D$ and $P$, and $B$ is between $A$ and $Q$). Now we chase angles.
$angle AQD = 180^{circ} - a-d$ and $angle DPC = a-d$ (since $angle DCP = 180^circ - a$).
Since $QH$ bisects $angle AQD$, we get $angle DHQ = 90 + frac a2 - frac d2 = $ and similarily $angle DFP = 180^circ - frac{a}{2} - frac d2$. That means that in the quadrilateral $DFIH$ we know three of the angles, and they add up to
$$
angle ADC + angle DFP + angle DHQ = d + 180^circ - frac{a}{2} - frac d2 + 90^circ + frac a{2} - frac d2 = 270^circ
$$
Thus the last angle in the quadrilateral must be $90^circ$.
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
$endgroup$
We need a few more letters: Let the intersection between the bisectors be at $I$, let $E, F$ be where the bisector at $P$ intersects $AB$ and $CD$ respectively, and let $G,H$ be where the bisector at $Q$ intersects $BC$ and $AD$ respectively. This gives the following figure:
(In retrospect, $E$ and $G$ were unnecessary, but hey, that's how you solve geometry puzzles: You keep adding points and lines and calculate until you find the ones you really needed.)
Let, for convinience, $angle DAB = a$ and $angle ADC = d$ (assuming the quadrilateral is named so that $A$ is between $D$ and $P$, and $B$ is between $A$ and $Q$). Now we chase angles.
$angle AQD = 180^{circ} - a-d$ and $angle DPC = a-d$ (since $angle DCP = 180^circ - a$).
Since $QH$ bisects $angle AQD$, we get $angle DHQ = 90 + frac a2 - frac d2 = $ and similarily $angle DFP = 180^circ - frac{a}{2} - frac d2$. That means that in the quadrilateral $DFIH$ we know three of the angles, and they add up to
$$
angle ADC + angle DFP + angle DHQ = d + 180^circ - frac{a}{2} - frac d2 + 90^circ + frac a{2} - frac d2 = 270^circ
$$
Thus the last angle in the quadrilateral must be $90^circ$.
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
answered Sep 14 '16 at 22:19
ArthurArthur
116k7116198
116k7116198
add a comment |
add a comment |
$begingroup$
Construction:- 1) Extend CP. 2) Draw PR, the external bisector of $angle APB$.
It should be clear that $theta = theta’$, $phi = phi’$, $omega = omega’$, and $n = n'$ [ext. angle, cyclic quad.]
The logic:- We know that the internal and the external angle bisector are perpendicular to each other. Thus, we are done if we can prove that PR // QI. The question now is changed to “will $omega = alpha?$”
By exterior angle, $omega = dfrac {m + n’}{2} = dfrac {(n + 2phi) + n'}{2} = n+phi= alpha$.
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
$endgroup$
add a comment |
$begingroup$
Construction:- 1) Extend CP. 2) Draw PR, the external bisector of $angle APB$.
It should be clear that $theta = theta’$, $phi = phi’$, $omega = omega’$, and $n = n'$ [ext. angle, cyclic quad.]
The logic:- We know that the internal and the external angle bisector are perpendicular to each other. Thus, we are done if we can prove that PR // QI. The question now is changed to “will $omega = alpha?$”
By exterior angle, $omega = dfrac {m + n’}{2} = dfrac {(n + 2phi) + n'}{2} = n+phi= alpha$.
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
$endgroup$
add a comment |
$begingroup$
Construction:- 1) Extend CP. 2) Draw PR, the external bisector of $angle APB$.
It should be clear that $theta = theta’$, $phi = phi’$, $omega = omega’$, and $n = n'$ [ext. angle, cyclic quad.]
The logic:- We know that the internal and the external angle bisector are perpendicular to each other. Thus, we are done if we can prove that PR // QI. The question now is changed to “will $omega = alpha?$”
By exterior angle, $omega = dfrac {m + n’}{2} = dfrac {(n + 2phi) + n'}{2} = n+phi= alpha$.
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
$endgroup$
Construction:- 1) Extend CP. 2) Draw PR, the external bisector of $angle APB$.
It should be clear that $theta = theta’$, $phi = phi’$, $omega = omega’$, and $n = n'$ [ext. angle, cyclic quad.]
The logic:- We know that the internal and the external angle bisector are perpendicular to each other. Thus, we are done if we can prove that PR // QI. The question now is changed to “will $omega = alpha?$”
By exterior angle, $omega = dfrac {m + n’}{2} = dfrac {(n + 2phi) + n'}{2} = n+phi= alpha$.
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
edited Dec 11 '18 at 16:21
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
answered Sep 15 '16 at 3:37
MickMick
11.9k21641
11.9k21641
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%2f1927069%2fproof-with-cyclic-quadrilateral-and-circle%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
