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5. In Fig.9.33, ABC and BDE are two equilateraltriangles such that D is the mid-point of BC. IfAEintersects BC at F, show that(i)ar(BDE)--ar(ABC)4Gi) ar (BDE)-ar(BAE)(Gii) ar (ABC)-2 ar (BEC)(iv) ar (BFE)- ar (AFD)(v) ar (BFE)-2 ar (FED)Fig. 9.33(vi) ar (FEDar (AFC |
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