triangles
Asked by | 27th Jan, 2009, 07:46: PM
Construction: Extend the median AD to a point E such that AD = DE
Extend the median PM to a point N such that PM = N
In DEC &
ABD
AD/DE= BD/DC ....................(Given)
ADB =
EDC..................(Vertically opposite angles)
DEC
ABD (BY SAS)..........(1)
Similarly, PQM
MNR..............(2)
DEC =
ABD &
QPM =
MNR (by cpct)
Now In AEC &
PNR
AB = DE & PQ = NR ( from eqn 1 & 2)
AC/PR = AE/PN= EC/NR
AEC
PNR (By SSS)
DAC =
MPR (cpct).........(3)
CED =
RNP (cpct)...........(4)
⇒ABD =
PQM
BAC =
QPR ( from 3 &4)
In ABC &
PQR
AB/PQ = AC/PR
ABC
PQR (By SAS)
Answered by | 5th Feb, 2009, 02:14: PM
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