Thursday, December 3, 2009

Problem 397: Triangle, Altitude, Midpoints, Congruence

Proposed Problem
Click the figure below to see the complete problem 397 about Triangle, Altitude, Midpoints, Congruence.

 Problem 397: Triangle, Altitude, Midpoints, Congruence.
See also:
Complete Problem 397
Level: High School, SAT Prep, College geometry


  1. DEF and DEH triangles have common side DE and DE//AC so have common heigth, then A(DEF)=A(DEH)

  2. Newzad,
    U've proven that the two triangles have the same area but not that they're congruent!

  3. Sorry, here is the proof of congruency

  4. suggest to others

    Prove DE is bisector of angle BDH (about congruence )

  5. 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.

  6. DEFH is cyclic.It is also a trapezium. Hence it must be isosceles trapezium. Hence proved.

  7. DE //AC. So < DEH = < EHC = <C = < FDE (since DECF is a parallelogram)
    = < 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
    Sri Lanka