Vikas TU
Last Activity: 5 Years ago
Dear student
Let DQ be the tangent to the circum-circle of Tr. ABD at D
. /_ADQ=/_CDQ = x.
Then by the Alternate Segment Theorem
/_DBA=x and likewise
/_BAD =x which makes
DA = DB = DC since D is the midpoint of BC.
In other words, D is the circum-centre of Tr. ABC and BC is its diameter. Then /_A = 90°.