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ΔDEF has DE=DF and O is a point on DE produced such that a circle with center O can be drawn to touch DF at F. If FE is produced cuts the circle at H, prove that OH is perpendicular to DE
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
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I'm not sure about the description and can't draw what you describe. Can you post the drawing?
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hi
Try this. Colour coded clue.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks Bob.
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thanks bob for the drawing.
here comes the proof:-
FD is the tangent to OF.
=> <OFD =90°
In ΔHOF,
<OHE= <OFE
In ΔDEF,
<DEF=<DFE
also, <DFE= 90° - <OFE
= 90° - <OHE
Now,
<HEO=<DEF
=> <HEO=90° - <OHE
=> <HEO +<OHE= 90°
Now, <HEO + <OHE + <HOE=180°
=> 90°+<HOE =180°
=> <HOE=90°
Hence, HO is perpendicular to OE.
proved.:)
friendship is tan 90°.
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Wow! Thanks Nihar
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
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you're most welcome
friendship is tan 90°.
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