Q. 2

# 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.

Answer :

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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