Asked by ashwinivg | 20th Mar, 2009, 07:45: AM
Construction: drop perpendiculars from D and B onto AC.Let these perpendiculars be DM and BN respectively..
consider are triangle DAE=1/2*(AE*DM)
area triangle BEC=1/2*(EC*BN)
area DAE* area BEC=1/4*(AE*DM*EC*BN)...(i)
area triangle AEB=1/2(AE*BN)
from (i) and (ii) we get,
the desired result.
Note that the perpendicular(height) for a triangle can be on the extended base as well like in the case of triangles DEC and AEB.
Answered by | 20th Mar, 2009, 08:46: AM
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