While practicing mathematics Olympiad questions , i got the below given question . Though the solution is given , I am not able to bypass certain steps ... Can anyone please explain me why angle KPA and angle KQA are each equal to 90-angle A ? I am not able to get this step . Awaiting for the get through ...
K is the orthocenter of ⊿APQ implies QKM is perpendicular to AP at M from Q.
Similarly, ⊿KNQ is right-angled at N. Therefore, MNQP and MKNA are cyclic quadrilaterals.
Hence, $\angle A = \angle MKP$ [ext. angle cyclic quad.]
$= 90^0 - \angle MPK$
That explains your query. If you need further help, say so.