In $triangle ABC$, with $D$ on $BC$, if perpendiculars from $D$ to the other sides are equal, can we conclude...
$begingroup$
In the figure, $triangle ABC$ is a triangle. There exists a point $D$ on $BC$ such that when two perpendicular lines are drawn from $D$ to $AB$ and $AC$, respectively, $DE = DF$, where $E$ and $F$ are the intersection points. Must it be true that $triangle ABC$ is isosceles?
It should be obvious to prove that $square AEDF$ is a kite, with $AE = AF$ and $DE=DF$. However, how can we draw any conclusion about $triangle ABC$? Any help/hint is appreciated.
geometry
edited Dec 11 '18 at 8:34
Blue
48.4k870154
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
$endgroup$
add a comment |
$begingroup$
In the figure, $triangle ABC$ is a triangle. There exists a point $D$ on $BC$ such that when two perpendicular lines are drawn from $D$ to $AB$ and $AC$, respectively, $DE = DF$, where $E$ and $F$ are the intersection points. Must it be true that $triangle ABC$ is isosceles?
It should be obvious to prove that $square AEDF$ is a kite, with $AE = AF$ and $DE=DF$. However, how can we draw any conclusion about $triangle ABC$? Any help/hint is appreciated.
geometry
edited Dec 11 '18 at 8:34
Blue
48.4k870154
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
$endgroup$
$begingroup$
You're correct about the kite. Now consider: Is there anything about the situation that guarantees the kite is balancing perfectly "upright"?
$endgroup$
– Blue
Dec 11 '18 at 8:29
$begingroup$
@Blue : How about rotating BC while fixing point D? Then I may not obtain an isosceles triangle?
$endgroup$
– LeeNL
Dec 11 '18 at 8:36
$begingroup$
Exactly. :) ${}{}$
$endgroup$
– Blue
Dec 11 '18 at 8:39
add a comment |
$begingroup$
In the figure, $triangle ABC$ is a triangle. There exists a point $D$ on $BC$ such that when two perpendicular lines are drawn from $D$ to $AB$ and $AC$, respectively, $DE = DF$, where $E$ and $F$ are the intersection points. Must it be true that $triangle ABC$ is isosceles?
It should be obvious to prove that $square AEDF$ is a kite, with $AE = AF$ and $DE=DF$. However, how can we draw any conclusion about $triangle ABC$? Any help/hint is appreciated.
geometry
edited Dec 11 '18 at 8:34
Blue
48.4k870154
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
$endgroup$
In the figure, $triangle ABC$ is a triangle. There exists a point $D$ on $BC$ such that when two perpendicular lines are drawn from $D$ to $AB$ and $AC$, respectively, $DE = DF$, where $E$ and $F$ are the intersection points. Must it be true that $triangle ABC$ is isosceles?
It should be obvious to prove that $square AEDF$ is a kite, with $AE = AF$ and $DE=DF$. However, how can we draw any conclusion about $triangle ABC$? Any help/hint is appreciated.
geometry
geometry
edited Dec 11 '18 at 8:34
Blue
48.4k870154
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
edited Dec 11 '18 at 8:34
Blue
48.4k870154
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
edited Dec 11 '18 at 8:34
Blue
48.4k870154
edited Dec 11 '18 at 8:34
Blue
48.4k870154
edited Dec 11 '18 at 8:34
Blue
48.4k870154
48.4k870154
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
asked Dec 11 '18 at 8:09
LeeNLLeeNL
9661129
9661129
$begingroup$
You're correct about the kite. Now consider: Is there anything about the situation that guarantees the kite is balancing perfectly "upright"?
$endgroup$
– Blue
Dec 11 '18 at 8:29
$begingroup$
@Blue : How about rotating BC while fixing point D? Then I may not obtain an isosceles triangle?
$endgroup$
– LeeNL
Dec 11 '18 at 8:36
$begingroup$
Exactly. :) ${}{}$
$endgroup$
– Blue
Dec 11 '18 at 8:39
add a comment |
$begingroup$
You're correct about the kite. Now consider: Is there anything about the situation that guarantees the kite is balancing perfectly "upright"?
$endgroup$
– Blue
Dec 11 '18 at 8:29
$begingroup$
@Blue : How about rotating BC while fixing point D? Then I may not obtain an isosceles triangle?
$endgroup$
– LeeNL
Dec 11 '18 at 8:36
$begingroup$
Exactly. :) ${}{}$
$endgroup$
– Blue
Dec 11 '18 at 8:39
$begingroup$
You're correct about the kite. Now consider: Is there anything about the situation that guarantees the kite is balancing perfectly "upright"?
$endgroup$
– Blue
Dec 11 '18 at 8:29
$begingroup$
You're correct about the kite. Now consider: Is there anything about the situation that guarantees the kite is balancing perfectly "upright"?
$endgroup$
– Blue
Dec 11 '18 at 8:29
$begingroup$
@Blue : How about rotating BC while fixing point D? Then I may not obtain an isosceles triangle?
$endgroup$
– LeeNL
Dec 11 '18 at 8:36
$begingroup$
@Blue : How about rotating BC while fixing point D? Then I may not obtain an isosceles triangle?
$endgroup$
– LeeNL
Dec 11 '18 at 8:36
$begingroup$
Exactly. :) ${}{}$
$endgroup$
– Blue
Dec 11 '18 at 8:39
$begingroup$
Exactly. :) ${}{}$
$endgroup$
– Blue
Dec 11 '18 at 8:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Join AD. Let angle $ADE = alpha$ and angle $ADF = beta$ then
$DE sec{alpha} =DF sec{beta} = AD$
(They both are equal to AD)
As $ DE = DF $, thus $alpha = beta$
Now $ADsin{alpha} = ADsin{beta}$
Thus $AE = AF$
Now we can see from above derived conditions that $Delta ABD$ is congruent with $Delta ACD$. Also that AD must be angle bisector of and at A.
But these conditions only fix A ,E, D and F (that they must form a cyclic quadrilateral). It doesn't tell anything about points B & C, not the queue at them.
To make it more intuitive, you can imagine that if we increase side AB a bit and reduce side AC a bit such that B, D and C remain collinear without disturbing AEFD ,triangle wouldn't be isosceles.
You can also use co-ordinate geometry to check further
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
$endgroup$
$begingroup$
Please check the congruent triangles.
$endgroup$
– LeeNL
Dec 15 '18 at 11:55
add a comment |
$begingroup$
All we can say is that $E$ and $F$ are points of a circle with center $D$, and $BC$ is on a line that passes thorough $D$.
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
$endgroup$
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%2f3035042%2fin-triangle-abc-with-d-on-bc-if-perpendiculars-from-d-to-the-other-si%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
$begingroup$
Join AD. Let angle $ADE = alpha$ and angle $ADF = beta$ then
$DE sec{alpha} =DF sec{beta} = AD$
(They both are equal to AD)
As $ DE = DF $, thus $alpha = beta$
Now $ADsin{alpha} = ADsin{beta}$
Thus $AE = AF$
Now we can see from above derived conditions that $Delta ABD$ is congruent with $Delta ACD$. Also that AD must be angle bisector of and at A.
But these conditions only fix A ,E, D and F (that they must form a cyclic quadrilateral). It doesn't tell anything about points B & C, not the queue at them.
To make it more intuitive, you can imagine that if we increase side AB a bit and reduce side AC a bit such that B, D and C remain collinear without disturbing AEFD ,triangle wouldn't be isosceles.
You can also use co-ordinate geometry to check further
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
$endgroup$
$begingroup$
Please check the congruent triangles.
$endgroup$
– LeeNL
Dec 15 '18 at 11:55
add a comment |
$begingroup$
Join AD. Let angle $ADE = alpha$ and angle $ADF = beta$ then
$DE sec{alpha} =DF sec{beta} = AD$
(They both are equal to AD)
As $ DE = DF $, thus $alpha = beta$
Now $ADsin{alpha} = ADsin{beta}$
Thus $AE = AF$
Now we can see from above derived conditions that $Delta ABD$ is congruent with $Delta ACD$. Also that AD must be angle bisector of and at A.
But these conditions only fix A ,E, D and F (that they must form a cyclic quadrilateral). It doesn't tell anything about points B & C, not the queue at them.
To make it more intuitive, you can imagine that if we increase side AB a bit and reduce side AC a bit such that B, D and C remain collinear without disturbing AEFD ,triangle wouldn't be isosceles.
You can also use co-ordinate geometry to check further
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
$endgroup$
$begingroup$
Please check the congruent triangles.
$endgroup$
– LeeNL
Dec 15 '18 at 11:55
add a comment |
$begingroup$
Join AD. Let angle $ADE = alpha$ and angle $ADF = beta$ then
$DE sec{alpha} =DF sec{beta} = AD$
(They both are equal to AD)
As $ DE = DF $, thus $alpha = beta$
Now $ADsin{alpha} = ADsin{beta}$
Thus $AE = AF$
Now we can see from above derived conditions that $Delta ABD$ is congruent with $Delta ACD$. Also that AD must be angle bisector of and at A.
But these conditions only fix A ,E, D and F (that they must form a cyclic quadrilateral). It doesn't tell anything about points B & C, not the queue at them.
To make it more intuitive, you can imagine that if we increase side AB a bit and reduce side AC a bit such that B, D and C remain collinear without disturbing AEFD ,triangle wouldn't be isosceles.
You can also use co-ordinate geometry to check further
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
$endgroup$
Join AD. Let angle $ADE = alpha$ and angle $ADF = beta$ then
$DE sec{alpha} =DF sec{beta} = AD$
(They both are equal to AD)
As $ DE = DF $, thus $alpha = beta$
Now $ADsin{alpha} = ADsin{beta}$
Thus $AE = AF$
Now we can see from above derived conditions that $Delta ABD$ is congruent with $Delta ACD$. Also that AD must be angle bisector of and at A.
But these conditions only fix A ,E, D and F (that they must form a cyclic quadrilateral). It doesn't tell anything about points B & C, not the queue at them.
To make it more intuitive, you can imagine that if we increase side AB a bit and reduce side AC a bit such that B, D and C remain collinear without disturbing AEFD ,triangle wouldn't be isosceles.
You can also use co-ordinate geometry to check further
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
answered Dec 11 '18 at 8:50
Abhik JainAbhik Jain
261
261
$begingroup$
Please check the congruent triangles.
$endgroup$
– LeeNL
Dec 15 '18 at 11:55
add a comment |
$begingroup$
Please check the congruent triangles.
$endgroup$
– LeeNL
Dec 15 '18 at 11:55
$begingroup$
Please check the congruent triangles.
$endgroup$
– LeeNL
Dec 15 '18 at 11:55
$begingroup$
Please check the congruent triangles.
$endgroup$
– LeeNL
Dec 15 '18 at 11:55
add a comment |
$begingroup$
All we can say is that $E$ and $F$ are points of a circle with center $D$, and $BC$ is on a line that passes thorough $D$.
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
$endgroup$
add a comment |
$begingroup$
All we can say is that $E$ and $F$ are points of a circle with center $D$, and $BC$ is on a line that passes thorough $D$.
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
$endgroup$
add a comment |
$begingroup$
All we can say is that $E$ and $F$ are points of a circle with center $D$, and $BC$ is on a line that passes thorough $D$.
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
$endgroup$
All we can say is that $E$ and $F$ are points of a circle with center $D$, and $BC$ is on a line that passes thorough $D$.
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
answered Dec 11 '18 at 9:29
Emilio NovatiEmilio Novati
52k43474
52k43474
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%2f3035042%2fin-triangle-abc-with-d-on-bc-if-perpendiculars-from-d-to-the-other-si%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
$begingroup$
You're correct about the kite. Now consider: Is there anything about the situation that guarantees the kite is balancing perfectly "upright"?
$endgroup$
– Blue
Dec 11 '18 at 8:29
$begingroup$
@Blue : How about rotating BC while fixing point D? Then I may not obtain an isosceles triangle?
$endgroup$
– LeeNL
Dec 11 '18 at 8:36
$begingroup$
Exactly. :) ${}{}$
$endgroup$
– Blue
Dec 11 '18 at 8:39