$O$ is the orthocentre of $triangle ABC$ if $APperp BC$, $BRperp AC$ and $CQ perp AB$. Prove that $angle OPQ= angle OPR$ [closed]
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$O$ is the orthocentre of $triangle ABC$ if $APperp BC$, $BRperp AC$ and $CQ perp AB$. Prove that $angle OPQ= angle OPR$
geometry euclidean-geometry
closed as off-topic by amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist Sep 10 at 0:11
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$O$ is the orthocentre of $triangle ABC$ if $APperp BC$, $BRperp AC$ and $CQ perp AB$. Prove that $angle OPQ= angle OPR$
geometry euclidean-geometry
closed as off-topic by amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist Sep 10 at 0:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist
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up vote
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$O$ is the orthocentre of $triangle ABC$ if $APperp BC$, $BRperp AC$ and $CQ perp AB$. Prove that $angle OPQ= angle OPR$
geometry euclidean-geometry
$O$ is the orthocentre of $triangle ABC$ if $APperp BC$, $BRperp AC$ and $CQ perp AB$. Prove that $angle OPQ= angle OPR$
geometry euclidean-geometry
geometry euclidean-geometry
edited Sep 9 at 12:54
Stefan4024
29.8k53377
29.8k53377
asked Sep 9 at 12:32
bhuwan budhathoki
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1
closed as off-topic by amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist Sep 10 at 0:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist
closed as off-topic by amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist Sep 10 at 0:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist
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As $APQC$ is cyclic we get that $angle BPQ = angle BAC$. Similarly as $ABPR$ is cyclic we get that $angle CPR = angle BAC$. Then we have:
$$angle APQ = angle BPA - angle BPQ = fracpi2 - angle BAC$$
$$angle APR = angle CPA - angle CPR = fracpi2 - angle BAC$$
Thus we get $angle APQ = angle APR$. Hence the proof.
Thank you soo much
â bhuwan budhathoki
Sep 9 at 13:25
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As $APQC$ is cyclic we get that $angle BPQ = angle BAC$. Similarly as $ABPR$ is cyclic we get that $angle CPR = angle BAC$. Then we have:
$$angle APQ = angle BPA - angle BPQ = fracpi2 - angle BAC$$
$$angle APR = angle CPA - angle CPR = fracpi2 - angle BAC$$
Thus we get $angle APQ = angle APR$. Hence the proof.
Thank you soo much
â bhuwan budhathoki
Sep 9 at 13:25
add a comment |Â
up vote
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As $APQC$ is cyclic we get that $angle BPQ = angle BAC$. Similarly as $ABPR$ is cyclic we get that $angle CPR = angle BAC$. Then we have:
$$angle APQ = angle BPA - angle BPQ = fracpi2 - angle BAC$$
$$angle APR = angle CPA - angle CPR = fracpi2 - angle BAC$$
Thus we get $angle APQ = angle APR$. Hence the proof.
Thank you soo much
â bhuwan budhathoki
Sep 9 at 13:25
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As $APQC$ is cyclic we get that $angle BPQ = angle BAC$. Similarly as $ABPR$ is cyclic we get that $angle CPR = angle BAC$. Then we have:
$$angle APQ = angle BPA - angle BPQ = fracpi2 - angle BAC$$
$$angle APR = angle CPA - angle CPR = fracpi2 - angle BAC$$
Thus we get $angle APQ = angle APR$. Hence the proof.
As $APQC$ is cyclic we get that $angle BPQ = angle BAC$. Similarly as $ABPR$ is cyclic we get that $angle CPR = angle BAC$. Then we have:
$$angle APQ = angle BPA - angle BPQ = fracpi2 - angle BAC$$
$$angle APR = angle CPA - angle CPR = fracpi2 - angle BAC$$
Thus we get $angle APQ = angle APR$. Hence the proof.
answered Sep 9 at 12:53
Stefan4024
29.8k53377
29.8k53377
Thank you soo much
â bhuwan budhathoki
Sep 9 at 13:25
add a comment |Â
Thank you soo much
â bhuwan budhathoki
Sep 9 at 13:25
Thank you soo much
â bhuwan budhathoki
Sep 9 at 13:25
Thank you soo much
â bhuwan budhathoki
Sep 9 at 13:25
add a comment |Â