1
ar (BDE)
ar (ABC)
4
1
ar (BAE)
2
(1) ar(BDE)
B
(1) ar
ar (ABC)=2 ar (BEC)
(iv) ar (BFE)= ar (AFD)
E
(vi) ar (FED)
ar (AFC)
8
(Hint: Join EC and AD. Show that BE || AC and DE IAB, etc.)
mangles such that D is the mid-point of BC IEAE
In Fig. 9.33, ABC and BDE are two equilateral
intersects BC at F. show that
(V) ar (BFE)= 2 ar (FED)
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