in triangle ABC, D and E are points on sides AB and AC respectively, such that AD:DB=5:4. DC and BE intersect at F.
Find ar(triangle DEF)/ar(triangle CBF
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Answers
Answer:
= 16/25
Step-by-step explanation:
DE || BC
AD/DB = AE/EF =4/5 ( Basic proportionality theorem)
In triangle ABE & ACD
AD/DB = AE/EC and angle A is common
So triangles ABE & ACD are similar
Therefore AD/DB = EF/FB ( properties of similar triangles)
Similarly AE / EC = BF/FC
In Triangles AEF & BFC
angle FEC is congruent to angle BFE ( alternate angles)
angel DEF is = angle BFC ( vertically opposite angles)
Therefore they are similar
Area of triangle DFE / area of BFC = (4/5)^2 ( Properties of similar triangles
= 16/25
heya..!!
here is ua answer:
AD/DB = AE/EF =4/5 ( Basic proportionality theorem)
In triangle ABE & ACD
AD/DB = AE/EC and angle A is common
So triangles ABE & ACD are similar
Therefore AD/DB = EF/FB ( properties of similar triangles)
Similarly AE / EC = BF/FC
In Triangles AEF & BFC
angle FEC is congruent to angle BFE ( alternate angles)
angel DEF is = angle BFC ( vertically opposite angles)
Therefore they are similar
Area of triangle DFE / area of BFC = (4/5)^2 ( Properties of similar triangles
= 16/25
hope it helps..!!♥