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Vedmedyk [2.9K]
3 years ago
12

In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

e ratio of BE:EC?
Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
8 0

Answer:

\frac{BE}{EC} =\frac{1}{3}

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

\frac{BK}{CK} =\frac{1}{4}

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE                ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE     ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK  and Δ EBK are similar

so we have

\frac{AD}{BE} =\frac{DK}{BK}

\frac{AD}{BE} =\frac{4}{1}

\frac{BC}{BE}=\frac{4}{1}     ( ABCD is parallelogram so AD=BC)

\frac{BE+EC}{BE}=\frac{4}{1}         ( BC= BE+EC)

\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}

1+\frac{EC}{BE}=4

\frac{EC}{BE}=3  ( subtracting 1 from both side )

\frac{EC}{BE}=\frac{3}{1}

taking reciprocal both side

\frac{BE}{EC} =\frac{1}{3}


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