In triangleABC seg DE is perpendicular to side BC If 2 A (ADE) =A(BDEC)Find AB/AD
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Answer:
BC = √3DE
Explanation:
In triangle ABC, seg DE || SEG BC. if twice area of triangle ADE = area of quadrilateral DBCE, find AB:AD. SHOW THAT BC = √3 d
DE || BC
hence ADE & ABC are similar
BC/DE = AB / AD = AC / AE = k
area of ABC = k^2 Area of ADE
Area of DBCE = Area of ABC - ARea of ADE
2(Area of ADE) = Area of DBCE
2(Area of ADE) = Area of ABC - ARea of ADE
3(Area of ADE) = Area of ABC
3(Area of ADE) = k^2 Area of ADE
k^2 = 3
k = √3
BC/DE = k = √3
BC = √3DE
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