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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. - Mathematics

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प्रश्न

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.

योग

उत्तर


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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अध्याय 8: Quadrilaterals - Exercise 8.4 [पृष्ठ ८२]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
अध्याय 8 Quadrilaterals
Exercise 8.4 | Q 2. | पृष्ठ ८२

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