$O$ 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







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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







    share|cite|improve this question















    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." – amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist
    If this question can be reworded to fit the rules in the help center, please edit the question.














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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







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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




      geometry euclidean-geometry






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      edited Sep 9 at 12:54









      Stefan4024

      29.8k53377




      29.8k53377










      asked Sep 9 at 12:32









      bhuwan budhathoki

      1




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      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." – amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist
      If this question can be reworded to fit the rules in the help center, please edit the question.




      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." – amWhy, Pierre-Guy Plamondon, Jyrki Lahtonen, Jendrik Stelzner, Theoretical Economist
      If this question can be reworded to fit the rules in the help center, please edit the question.




















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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.






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          • Thank you soo much
            – bhuwan budhathoki
            Sep 9 at 13:25

















          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

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          active

          oldest

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          up vote
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          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.






          share|cite|improve this answer




















          • Thank you soo much
            – bhuwan budhathoki
            Sep 9 at 13:25














          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.






          share|cite|improve this answer




















          • Thank you soo much
            – bhuwan budhathoki
            Sep 9 at 13:25












          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 9 at 12:53









          Stefan4024

          29.8k53377




          29.8k53377











          • Thank you soo much
            – bhuwan budhathoki
            Sep 9 at 13:25
















          • Thank you soo much
            – bhuwan budhathoki
            Sep 9 at 13:25















          Thank you soo much
          – bhuwan budhathoki
          Sep 9 at 13:25




          Thank you soo much
          – bhuwan budhathoki
          Sep 9 at 13:25


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