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In a parallelogram ABCD, AB = 10 cm and ...

In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of `/_`A meets DC in E. AE and BC produced meet at F. Find the length of CF.

Text Solution

Verified by Experts

`/_ADE` is a isosceles triangle
AD=DE=6cm
EC=DC-DE
=10-6=4cm
`/_EFC`~`/_AFB`
`(AB)/(EC)=(BF)/(CF)`
`10/4=6/(CF)+1`
`CF=4cm.
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