Question

In parallelogram ABCD , AB=10cm, AD=6cm. The bisector of angle A meets BC in E . AE and BC priduced and meet at F. Find the length of CF

Answer 1

Here AB = 10 cm, AD = 6 cm

∴ DC = AB = 10 cm

AD = BC = 6 cm

Bisector of ∠A meets DE on E.

Produce AE and BC to meet at point F. Extend AD to G. From F draw HF || CD.

To find: CF

We have CD || FH and CF || DH .

So, DCFH is parallelogram

Also. AB || FH and AH || BF.

⇒ ABFH is also parallelogram.

Consider ΔAHF and ΔABF :

∠AHF = ∠ABF [Opposite angles are equal]

AF = AF (Common side)

and ∠HAF = ∠FAB

ΔAFH is congruent ΔABF (by AAS congruency)

⇒ AB = AH (CPCT)

⇒ AB = AH = AD + DH = AD + CF [DCFH is a parallelogram]

∴ CF = AB – AD = (10 – 6) cm= 4 cm

Was this answer helpful????? if it is so mark as brainliest..

Answer 2

HEY FRIEND HERE YOUR ANSWER

"NOTE"

V.O.C= vertically opposite angle

BCOZ= because

CF=4 cm
my handwriting so bad yaar

I HOPE IT'S HELP YOU

MARK BRAINIEST MY ANSWER