Proposed Problem
Click the figure below to see the complete problem 397 about Triangle, Altitude, Midpoints, Congruence.
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Complete Problem 397
Level: High School, SAT Prep, College geometry
Thursday, December 3, 2009
Problem 397: Triangle, Altitude, Midpoints, Congruence
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DEF and DEH triangles have common side DE and DE//AC so have common heigth, then A(DEF)=A(DEH)
ReplyDeleteNewzad,
ReplyDeleteU've proven that the two triangles have the same area but not that they're congruent!
Ajit
Sorry, here is the proof of congruency
ReplyDeletehttp://i48.tinypic.com/nx2l1v.gif
suggest to others
ReplyDeleteProve DE is bisector of angle BDH (about congruence )
Triangle AHB is right angled and D is midpoint of AB. So, AD=DB=DH. So angle DAH = angle DHA. Also AD parallel to EF and AD=EF (as EF = (1/2)AB = AD). So DEFA is a parallellogram. So, DH = DA = EF. Now DE parallel to FH. So angle EDH = DHA = DAH. Now angle DAH = angle DEF. Therefore, angle EDH = angle DEF. In triangles DHE and DEF, DH = FE, DE common, angle EDH = DEF. So the proof follows according to S-A-S rule of congruence.
ReplyDeleteDEFH is cyclic.It is also a trapezium. Hence it must be isosceles trapezium. Hence proved.
ReplyDeleteDE //AC. So < DEH = < EHC = <C = < FDE (since DECF is a parallelogram)
ReplyDelete= < DFH ….(1)
From (1), DEFH is concyclic, so < DHE = < DFE…..(2)
Hence, in Tr.s DHE and DFE,
DE is common,
< DEH = < FDE from (1) and
<DHE = < DFE from (2)
So, Tr.s DHE ≡ DFE (ASA)
Sumith Peiris
Moratuwa
Sri Lanka