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Question
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.
Solution
Given, AB = 10 cm, AD = 6 cm
DC = AB = 10 cm and AD = BC = 6 cm
Given, bisector of ∠A intersects DE at E and BC produced at F.
Now, drawing PF || CD.
From the figure,
CD || FP and CF || DP
PDCF is a parallelogram.
And AB || FP and AP || BF
ABFP is also a parallelogram
In ΔAPF and ΔABF
∠APF = ∠ABF ...(Opposite angles of a parallelogram are equal)
AF = AF ...(Common side)
∠PAF = ∠AFB ...(Alternate angles)
ΔAPF ≅ ΔABF ...(By ASA congruence criterion)
AB = AP ...(CPCT)
AB = AD + DP
= AD + CF ...(Since DCFP is a parallelogram)
∴ CF = AB – AD
= (10 – 6) cm
= 4 cm
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