Let $ABCD$ be a parallelogram. Show that $angle BQD = 90^ circ$.
$begingroup$
Let $ABCD$ be a parallelogram. There is point $P$ on the $BD$ such that $AP=BD$. $Q$ is the midpoint of the $CP$. Show that $angle BQD = 90^ circ$.
geometry euclidean-geometry geometric-transformation
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
$endgroup$
add a comment |
$begingroup$
Let $ABCD$ be a parallelogram. There is point $P$ on the $BD$ such that $AP=BD$. $Q$ is the midpoint of the $CP$. Show that $angle BQD = 90^ circ$.
geometry euclidean-geometry geometric-transformation
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
$endgroup$
add a comment |
$begingroup$
Let $ABCD$ be a parallelogram. There is point $P$ on the $BD$ such that $AP=BD$. $Q$ is the midpoint of the $CP$. Show that $angle BQD = 90^ circ$.
geometry euclidean-geometry geometric-transformation
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
$endgroup$
Let $ABCD$ be a parallelogram. There is point $P$ on the $BD$ such that $AP=BD$. $Q$ is the midpoint of the $CP$. Show that $angle BQD = 90^ circ$.
geometry euclidean-geometry geometric-transformation
geometry euclidean-geometry geometric-transformation
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
edited Dec 25 '18 at 17:17
Maria Mazur
46.6k1260119
46.6k1260119
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
asked Sep 1 '18 at 8:13
math.troublemath.trouble
546
546
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Alternatively, draw the second diagonal and denote the center with $O$.
$hspace{4cm}$
Using the formulas of medians:
$$begin{align}DQ^2&=frac{2CD^2+2PD^2-PC^2}{4};\
BQ^2&=frac{2BC^2+2BP^2-PC^2}{4};\
OP^2&=frac{2AP^2+2PC^2-AC^2}{4} Rightarrow color{red}{AC^2}=2AP^2+2PC^2-4OP^2=color{red}{2BD^2+2PC^2-4OP^2}.end{align}$$
Adding:
$$begin{align}DQ^2+BQ^2&=frac{(2CD^2+2BC^2)+2(PD^2+BP^2)-2PC^2}{4}=\
&=frac{(color{red}{AC^2}+BD^2)+2left(left(frac{BD}{2}+OPright)^2+left(frac{BD}{2}-OPright)^2right)-2PC^2}{4}=\
&=frac{(color{red}{2BD^2+2PC^2-4OP^2}+BD^2)+2left(frac{BD^2}{2}+2OP^2right)-2PC^2}{4}=\
&=BD^2.end{align}$$
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
Translate $P$ by $vec{AD}$ into new point $R$. Then triangle $BDR$ is isosceles and $BCRP$ is a parallelogram. Since $Q$ is a midpoint of a diagonal of parallelogram $BCRP$ it halves $BR$ also. But then $DQ$ is an altitude in isosceles triangle $BDR$ and thus a conclusion.
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
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%2f2901486%2flet-abcd-be-a-parallelogram-show-that-angle-bqd-90-circ%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
$begingroup$
Alternatively, draw the second diagonal and denote the center with $O$.
$hspace{4cm}$
Using the formulas of medians:
$$begin{align}DQ^2&=frac{2CD^2+2PD^2-PC^2}{4};\
BQ^2&=frac{2BC^2+2BP^2-PC^2}{4};\
OP^2&=frac{2AP^2+2PC^2-AC^2}{4} Rightarrow color{red}{AC^2}=2AP^2+2PC^2-4OP^2=color{red}{2BD^2+2PC^2-4OP^2}.end{align}$$
Adding:
$$begin{align}DQ^2+BQ^2&=frac{(2CD^2+2BC^2)+2(PD^2+BP^2)-2PC^2}{4}=\
&=frac{(color{red}{AC^2}+BD^2)+2left(left(frac{BD}{2}+OPright)^2+left(frac{BD}{2}-OPright)^2right)-2PC^2}{4}=\
&=frac{(color{red}{2BD^2+2PC^2-4OP^2}+BD^2)+2left(frac{BD^2}{2}+2OP^2right)-2PC^2}{4}=\
&=BD^2.end{align}$$
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
Alternatively, draw the second diagonal and denote the center with $O$.
$hspace{4cm}$
Using the formulas of medians:
$$begin{align}DQ^2&=frac{2CD^2+2PD^2-PC^2}{4};\
BQ^2&=frac{2BC^2+2BP^2-PC^2}{4};\
OP^2&=frac{2AP^2+2PC^2-AC^2}{4} Rightarrow color{red}{AC^2}=2AP^2+2PC^2-4OP^2=color{red}{2BD^2+2PC^2-4OP^2}.end{align}$$
Adding:
$$begin{align}DQ^2+BQ^2&=frac{(2CD^2+2BC^2)+2(PD^2+BP^2)-2PC^2}{4}=\
&=frac{(color{red}{AC^2}+BD^2)+2left(left(frac{BD}{2}+OPright)^2+left(frac{BD}{2}-OPright)^2right)-2PC^2}{4}=\
&=frac{(color{red}{2BD^2+2PC^2-4OP^2}+BD^2)+2left(frac{BD^2}{2}+2OP^2right)-2PC^2}{4}=\
&=BD^2.end{align}$$
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
Alternatively, draw the second diagonal and denote the center with $O$.
$hspace{4cm}$
Using the formulas of medians:
$$begin{align}DQ^2&=frac{2CD^2+2PD^2-PC^2}{4};\
BQ^2&=frac{2BC^2+2BP^2-PC^2}{4};\
OP^2&=frac{2AP^2+2PC^2-AC^2}{4} Rightarrow color{red}{AC^2}=2AP^2+2PC^2-4OP^2=color{red}{2BD^2+2PC^2-4OP^2}.end{align}$$
Adding:
$$begin{align}DQ^2+BQ^2&=frac{(2CD^2+2BC^2)+2(PD^2+BP^2)-2PC^2}{4}=\
&=frac{(color{red}{AC^2}+BD^2)+2left(left(frac{BD}{2}+OPright)^2+left(frac{BD}{2}-OPright)^2right)-2PC^2}{4}=\
&=frac{(color{red}{2BD^2+2PC^2-4OP^2}+BD^2)+2left(frac{BD^2}{2}+2OP^2right)-2PC^2}{4}=\
&=BD^2.end{align}$$
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
$endgroup$
Alternatively, draw the second diagonal and denote the center with $O$.
$hspace{4cm}$
Using the formulas of medians:
$$begin{align}DQ^2&=frac{2CD^2+2PD^2-PC^2}{4};\
BQ^2&=frac{2BC^2+2BP^2-PC^2}{4};\
OP^2&=frac{2AP^2+2PC^2-AC^2}{4} Rightarrow color{red}{AC^2}=2AP^2+2PC^2-4OP^2=color{red}{2BD^2+2PC^2-4OP^2}.end{align}$$
Adding:
$$begin{align}DQ^2+BQ^2&=frac{(2CD^2+2BC^2)+2(PD^2+BP^2)-2PC^2}{4}=\
&=frac{(color{red}{AC^2}+BD^2)+2left(left(frac{BD}{2}+OPright)^2+left(frac{BD}{2}-OPright)^2right)-2PC^2}{4}=\
&=frac{(color{red}{2BD^2+2PC^2-4OP^2}+BD^2)+2left(frac{BD^2}{2}+2OP^2right)-2PC^2}{4}=\
&=BD^2.end{align}$$
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
answered Sep 1 '18 at 10:29
farruhotafarruhota
20.8k2741
20.8k2741
add a comment |
add a comment |
$begingroup$
Translate $P$ by $vec{AD}$ into new point $R$. Then triangle $BDR$ is isosceles and $BCRP$ is a parallelogram. Since $Q$ is a midpoint of a diagonal of parallelogram $BCRP$ it halves $BR$ also. But then $DQ$ is an altitude in isosceles triangle $BDR$ and thus a conclusion.
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
add a comment |
$begingroup$
Translate $P$ by $vec{AD}$ into new point $R$. Then triangle $BDR$ is isosceles and $BCRP$ is a parallelogram. Since $Q$ is a midpoint of a diagonal of parallelogram $BCRP$ it halves $BR$ also. But then $DQ$ is an altitude in isosceles triangle $BDR$ and thus a conclusion.
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
add a comment |
$begingroup$
Translate $P$ by $vec{AD}$ into new point $R$. Then triangle $BDR$ is isosceles and $BCRP$ is a parallelogram. Since $Q$ is a midpoint of a diagonal of parallelogram $BCRP$ it halves $BR$ also. But then $DQ$ is an altitude in isosceles triangle $BDR$ and thus a conclusion.
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
Translate $P$ by $vec{AD}$ into new point $R$. Then triangle $BDR$ is isosceles and $BCRP$ is a parallelogram. Since $Q$ is a midpoint of a diagonal of parallelogram $BCRP$ it halves $BR$ also. But then $DQ$ is an altitude in isosceles triangle $BDR$ and thus a conclusion.
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
edited Sep 1 '18 at 8:39
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
answered Sep 1 '18 at 8:31
Maria MazurMaria Mazur
46.6k1260119
46.6k1260119
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%2f2901486%2flet-abcd-be-a-parallelogram-show-that-angle-bqd-90-circ%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
