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, ABCD is a parallelogram
AB = 10 cm
AD = 6 cm
The bisector of ∠A meets DC in E.
AE and BC are produced to meet at F.
We have to find the length of CF.
Now extend AD to H and join HF.
So, ABFH is a parallelogram
We know that the opposite sides of a parallelogram are parallel and congruent.
AB || HF and HF = AB
We know that the alternate interior angles are equal.
∠AFH = ∠FAB ------------------------- (1)
Since AF is the bisector of ∠A
∠HAF = ∠FAB ------------------------ (2)
From (1) and (2),
∠AFH = ∠HAF
We know that the sides opposite to equal angles are equal.
HF = AH
Since, HF = AB
HF = 10 cm
Since HF = AH, AH = 10 cm
From the figure,
AH = AD + DH
10 = 6 + DH
DH = 10 - 6
DH = 4 cm
CFHD is a parallelogram
We know that the opposite sides of a parallelogram are parallel and congruent.
So, DH = CF
Therefore, CF = 4cm
✦ Try This: Show that in a right angled triangle, the hypotenuse is the longest side.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8
NCERT Exemplar Class 9 Maths Exercise 8.4 Problem 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.
Summary:
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. The length of CF is 4 cm
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