How to prove that the given lines are perpendicular?
$begingroup$
I have a triangle $triangle ABC$, with circumcentre $O$, the reflection of $A$ through $BC$ gives me $T$. The perpendicular bisector of $TC$ and $TB$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively.
I need to prove that $TO$ is perpendicular to $PQ$.
How do I prove it? I tried some simple angle chasing but couldn't come to any significant conclusion. The solution booklet however tells me to use the fact that the reflection of $O$ in $BC$, say $O'$ lies on the circumcircle of $triangle APQ$, how did he arrive at this conclusion?
geometry
edited Sep 18 '18 at 9:36
limeeattack
727
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
$endgroup$
add a comment |
$begingroup$
I have a triangle $triangle ABC$, with circumcentre $O$, the reflection of $A$ through $BC$ gives me $T$. The perpendicular bisector of $TC$ and $TB$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively.
I need to prove that $TO$ is perpendicular to $PQ$.
How do I prove it? I tried some simple angle chasing but couldn't come to any significant conclusion. The solution booklet however tells me to use the fact that the reflection of $O$ in $BC$, say $O'$ lies on the circumcircle of $triangle APQ$, how did he arrive at this conclusion?
geometry
edited Sep 18 '18 at 9:36
limeeattack
727
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
$endgroup$
add a comment |
$begingroup$
I have a triangle $triangle ABC$, with circumcentre $O$, the reflection of $A$ through $BC$ gives me $T$. The perpendicular bisector of $TC$ and $TB$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively.
I need to prove that $TO$ is perpendicular to $PQ$.
How do I prove it? I tried some simple angle chasing but couldn't come to any significant conclusion. The solution booklet however tells me to use the fact that the reflection of $O$ in $BC$, say $O'$ lies on the circumcircle of $triangle APQ$, how did he arrive at this conclusion?
geometry
edited Sep 18 '18 at 9:36
limeeattack
727
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
$endgroup$
I have a triangle $triangle ABC$, with circumcentre $O$, the reflection of $A$ through $BC$ gives me $T$. The perpendicular bisector of $TC$ and $TB$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively.
I need to prove that $TO$ is perpendicular to $PQ$.
How do I prove it? I tried some simple angle chasing but couldn't come to any significant conclusion. The solution booklet however tells me to use the fact that the reflection of $O$ in $BC$, say $O'$ lies on the circumcircle of $triangle APQ$, how did he arrive at this conclusion?
geometry
geometry
edited Sep 18 '18 at 9:36
limeeattack
727
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
edited Sep 18 '18 at 9:36
limeeattack
727
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
edited Sep 18 '18 at 9:36
limeeattack
727
edited Sep 18 '18 at 9:36
limeeattack
727
edited Sep 18 '18 at 9:36
limeeattack
727
727
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
asked Sep 18 '18 at 6:33
saisanjeevsaisanjeev
1,025312
1,025312
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $PR$ be the perpendicular bisector of side $TC$ , and $SQ$ be the perpendicular bisector of side $BT$ . Let them intersect at point $O’$.
Observe that $O’$ is the circumcentre of $triangle BCT$ . It is thus the reflection of $O$ in $BC$.
$$∠RPA + ∠SQA = ∠O’PA+O’QA = (360-90-2∠C-A)+(360-90-2∠B-A)
$$ $$=720-180-2(∠A+∠B+∠C) = 720 - 540 = 180 .$$
Thus , the angles are supplementary. This implies $O’PAQ$ is a cyclic quadrilateral, and the points are concyclic.
Join the diagonals and $OO’$ . Join $O’A$ and $OT$, to meet at $G$. Let $OO’$ meet $BC$ at $D$ . Let $AT$ meet $BC$ at $H$ .
For convenience , Let $∠BAH = w , ∠HAG = y , ∠GAQ = x$ .
Using the properties of cyclic quadrilaterals and by angle chasing , we get $∠BMP = y .$ Also observe that $ ∠HAG = ∠HTG = y$ and since $AT$ is parallel to $OO’ , ∠TOO’ = y.$
Then , the line segments $PQ$ and $OT$ are inclined to $BC$ and $OO’$ respectively at an equal angle . This implies that the rotation of these line segments by that angle , will not change the angle between them , and the angle between them will equal $angle ODB$. But $∠ODB = 90.$
Hence , OT is perpendicular to PQ
answered Sep 29 '18 at 18:47
SinπSinπ
64511
$endgroup$
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%2f2921160%2fhow-to-prove-that-the-given-lines-are-perpendicular%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $PR$ be the perpendicular bisector of side $TC$ , and $SQ$ be the perpendicular bisector of side $BT$ . Let them intersect at point $O’$.
Observe that $O’$ is the circumcentre of $triangle BCT$ . It is thus the reflection of $O$ in $BC$.
$$∠RPA + ∠SQA = ∠O’PA+O’QA = (360-90-2∠C-A)+(360-90-2∠B-A)
$$ $$=720-180-2(∠A+∠B+∠C) = 720 - 540 = 180 .$$
Thus , the angles are supplementary. This implies $O’PAQ$ is a cyclic quadrilateral, and the points are concyclic.
Join the diagonals and $OO’$ . Join $O’A$ and $OT$, to meet at $G$. Let $OO’$ meet $BC$ at $D$ . Let $AT$ meet $BC$ at $H$ .
For convenience , Let $∠BAH = w , ∠HAG = y , ∠GAQ = x$ .
Using the properties of cyclic quadrilaterals and by angle chasing , we get $∠BMP = y .$ Also observe that $ ∠HAG = ∠HTG = y$ and since $AT$ is parallel to $OO’ , ∠TOO’ = y.$
Then , the line segments $PQ$ and $OT$ are inclined to $BC$ and $OO’$ respectively at an equal angle . This implies that the rotation of these line segments by that angle , will not change the angle between them , and the angle between them will equal $angle ODB$. But $∠ODB = 90.$
Hence , OT is perpendicular to PQ
answered Sep 29 '18 at 18:47
SinπSinπ
64511
$endgroup$
add a comment |
$begingroup$
Let $PR$ be the perpendicular bisector of side $TC$ , and $SQ$ be the perpendicular bisector of side $BT$ . Let them intersect at point $O’$.
Observe that $O’$ is the circumcentre of $triangle BCT$ . It is thus the reflection of $O$ in $BC$.
$$∠RPA + ∠SQA = ∠O’PA+O’QA = (360-90-2∠C-A)+(360-90-2∠B-A)
$$ $$=720-180-2(∠A+∠B+∠C) = 720 - 540 = 180 .$$
Thus , the angles are supplementary. This implies $O’PAQ$ is a cyclic quadrilateral, and the points are concyclic.
Join the diagonals and $OO’$ . Join $O’A$ and $OT$, to meet at $G$. Let $OO’$ meet $BC$ at $D$ . Let $AT$ meet $BC$ at $H$ .
For convenience , Let $∠BAH = w , ∠HAG = y , ∠GAQ = x$ .
Using the properties of cyclic quadrilaterals and by angle chasing , we get $∠BMP = y .$ Also observe that $ ∠HAG = ∠HTG = y$ and since $AT$ is parallel to $OO’ , ∠TOO’ = y.$
Then , the line segments $PQ$ and $OT$ are inclined to $BC$ and $OO’$ respectively at an equal angle . This implies that the rotation of these line segments by that angle , will not change the angle between them , and the angle between them will equal $angle ODB$. But $∠ODB = 90.$
Hence , OT is perpendicular to PQ
answered Sep 29 '18 at 18:47
SinπSinπ
64511
$endgroup$
add a comment |
$begingroup$
Let $PR$ be the perpendicular bisector of side $TC$ , and $SQ$ be the perpendicular bisector of side $BT$ . Let them intersect at point $O’$.
Observe that $O’$ is the circumcentre of $triangle BCT$ . It is thus the reflection of $O$ in $BC$.
$$∠RPA + ∠SQA = ∠O’PA+O’QA = (360-90-2∠C-A)+(360-90-2∠B-A)
$$ $$=720-180-2(∠A+∠B+∠C) = 720 - 540 = 180 .$$
Thus , the angles are supplementary. This implies $O’PAQ$ is a cyclic quadrilateral, and the points are concyclic.
Join the diagonals and $OO’$ . Join $O’A$ and $OT$, to meet at $G$. Let $OO’$ meet $BC$ at $D$ . Let $AT$ meet $BC$ at $H$ .
For convenience , Let $∠BAH = w , ∠HAG = y , ∠GAQ = x$ .
Using the properties of cyclic quadrilaterals and by angle chasing , we get $∠BMP = y .$ Also observe that $ ∠HAG = ∠HTG = y$ and since $AT$ is parallel to $OO’ , ∠TOO’ = y.$
Then , the line segments $PQ$ and $OT$ are inclined to $BC$ and $OO’$ respectively at an equal angle . This implies that the rotation of these line segments by that angle , will not change the angle between them , and the angle between them will equal $angle ODB$. But $∠ODB = 90.$
Hence , OT is perpendicular to PQ
answered Sep 29 '18 at 18:47
SinπSinπ
64511
$endgroup$
Let $PR$ be the perpendicular bisector of side $TC$ , and $SQ$ be the perpendicular bisector of side $BT$ . Let them intersect at point $O’$.
Observe that $O’$ is the circumcentre of $triangle BCT$ . It is thus the reflection of $O$ in $BC$.
$$∠RPA + ∠SQA = ∠O’PA+O’QA = (360-90-2∠C-A)+(360-90-2∠B-A)
$$ $$=720-180-2(∠A+∠B+∠C) = 720 - 540 = 180 .$$
Thus , the angles are supplementary. This implies $O’PAQ$ is a cyclic quadrilateral, and the points are concyclic.
Join the diagonals and $OO’$ . Join $O’A$ and $OT$, to meet at $G$. Let $OO’$ meet $BC$ at $D$ . Let $AT$ meet $BC$ at $H$ .
For convenience , Let $∠BAH = w , ∠HAG = y , ∠GAQ = x$ .
Using the properties of cyclic quadrilaterals and by angle chasing , we get $∠BMP = y .$ Also observe that $ ∠HAG = ∠HTG = y$ and since $AT$ is parallel to $OO’ , ∠TOO’ = y.$
Then , the line segments $PQ$ and $OT$ are inclined to $BC$ and $OO’$ respectively at an equal angle . This implies that the rotation of these line segments by that angle , will not change the angle between them , and the angle between them will equal $angle ODB$. But $∠ODB = 90.$
Hence , OT is perpendicular to PQ
answered Sep 29 '18 at 18:47
SinπSinπ
64511
edited Dec 30 '18 at 9:13
answered Sep 29 '18 at 18:47
SinπSinπ
64511
answered Sep 29 '18 at 18:47
SinπSinπ
64511
answered Sep 29 '18 at 18:47
SinπSinπ
64511
64511
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%2f2921160%2fhow-to-prove-that-the-given-lines-are-perpendicular%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
