Vikas TU
Last Activity: 7 Years ago
Dear Student,
Though it is hard to prove without diagram:-
H' = refection of H using line AB
D is the second end of CD.
ang CBD = 90 (as an inscribed angle subtended by a diameter),
=>BD perpendicular to BC. AH is an altitude in ABC, is perpendicular to BC
so AH parallel to BD.
BH is an altitude while ang CAD = 90. Therefore AD parallel to BH.
ADBH is a parallelogram. AB and DH intersect at common midpoint so midpoint Mc of AB lies on DH
and HMc = McD.
C' be the foot of the altitude CH HC' = C'H'.
In triangle DHH' C'Mc is parallel to DH': C'Mc parallel toDH'. And, since ang BC'C = 90, so too ang DH'H = 90. ang DH'C =90 and subtended by CD.
so, H' lies on the circumcircle. [proved]
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
