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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.

**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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