## 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.

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)

U've proven that the two triangles have the same area but not that they're congruent!
Ajit

3. Sorry, here is the proof of congruency

http://i48.tinypic.com/nx2l1v.gif

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
Moratuwa
Sri Lanka