I am not super sure but....let assume H is not on gamma 3, then it would be in interior...extend PH to meet gamma 3 at H'. Now, APH'T is cyclic...also, angle PAT + angle PHT is 180°( in quad. APHT, .....as ang. PAB= ang. BPM and ang. TAB= ang. BTM...then using the fact about the congruent triag.. PBT and PHT). So, it implies angle at H and H' are same....hence H coincides with H'. Please correct me if i am wrong....just did it with the diagram in the video..... PS- Tbh i hardly did these types of problems in school. Wish had such exposure at that time....this is how maths should be taught and learnt.
I am not super sure but....let assume H is not on gamma 3, then it would be in interior...extend PH to meet gamma 3 at H'. Now, APH'T is cyclic...also, angle PAT + angle PHT is 180°( in quad. APHT, .....as ang. PAB= ang. BPM and ang. TAB= ang. BTM...then using the fact about the congruent triag.. PBT and PHT). So, it implies angle at H and H' are same....hence H coincides with H'.
Please correct me if i am wrong....just did it with the diagram in the video.....
PS- Tbh i hardly did these types of problems in school. Wish had such exposure at that time....this is how maths should be taught and learnt.