Answers

2016-01-01T11:16:56+05:30
GIVEN- 
ar triangle ABC= ar triangle DBC
They have same base BC and have same area.
TPT-
BC bisects AD
construction-
let AO be the height of triangle ABC and let DO be the height of triangle DBC.
PROOF-
In triangle ABC and DBC
They have same bases and equal area
that is 1/2 X BC X AO =1/2 X BC X DO
As area triangle ABC = ar triangle DBC
Therefore AO=DO
Therefore BC bisects AD.
Hence proved..

THANKS FOR ASKING

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2016-01-01T19:11:35+05:30
let ABC and DBC be two triangles with AE passes through BC to D
Now,  ar (Δ ABC) = ar (ΔDBC)
             1/2 * AE * BC = 1/2 * DE * BC
           so, AE = DE
∴ BC bisects AD
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