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

Labels:
altitude,
congruence,
midpoint,
triangle

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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