Question

In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of $\angle A$ meets DC in E. AE and BC produced meet at F. The length of CF is __ cm.

Answer 1

Given, a parallelogram ABCD in which AB = 10 cm and AD = 6cm.

Now, draw a bisector of $\angle A$ meets DC in E and produce it to F and produce BC to meet at F.

Also, produce AD to H and join HF, so that ABFH is a parallelogram.

Since HF ∥ AB

$\therefore \angle AFH=\angle FAB$ [alternate interior angles]

$\angle HAF=\angle FAB$ [since, AF is the bisector of $\angle A$]

$\Rightarrow \angle HAF=\angle AFH$ [from eq.(i)]

$\Rightarrow HF=AH$ [sides opposite to equal angles are equal ]

But HF = AB = 10 cm

$\therefore AH=AB=10cm$

$\Rightarrow AD+DH=10cm$

$\Rightarrow DH=(10-6)cm$

$\therefore DH=4cm$

Since CFHD is a parallelogram.

Therefore, the opposite sides are equal.

$\therefore DH=CF=4cm$