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Pace_T

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see attachment
the one circled in red
thx
 

word.

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I like to complicate questions, so:

Let T{ABC} denote "Triangle ABC" and A{ABC} "Angle ABC".

You can prove T{BAG} is congruent to T{ADE} through SAS.

Construct a line from M through A to point X on BG.

T{AMD} is isosceles (AM = DM, equal radii, part)
hence A{MAD} = A{ADM} (base angles of isos.T{AMD})
Also, A{XBA} = A{ADM} (corresponding angles in congruent triangles)
Therefore A{XBA} = A{MAD}

A{BXA} + A{XBA} = A{MAD} + 90° (exterior angle of T{BAX})
Therefore A{BXA} = 90°
 
Last edited:

Riviet

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*applauds for who loves maths* Very nice use of trig! :)
 

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