Question

ABCD is a parallelogram. If E is the midpoint of BC, and AE the bisector of $\angle A$ , prove that $AB=\frac{1}{2}AD$

Answer 1

Since AE is the bisector of $\angle A$.
Therefore
$\angle 1=\frac{1}{2}\angle A$ . . . . .. (1)
Since ABCD is a parallelogram is parallelogram
Therefore, AD || BC and AB intersects them.
$\Rightarrow \angle A+\angle B=180$ (sum of interior angles is 180)
$\Rightarrow \angle B=180+\angle A$
In $△ABE$, we have
$\angle 1+\angle 2+\angle B=180$
$\Rightarrow \frac{1}{2}\angle A+\angle 2+180-\angle A=180$
$\angle 2-\frac{1}{2}\angle A=0$
$\angle 2=\frac{1}{2}\angle A$ . . . . . (2)
from (1) and (2)
$\angle 1=\angle 2$
In $△ABE$ we have
BE = AB (sides opposite to equal angles are equal)
2 BE = 2 AB (multiplying by 2 both sides)
BC = 2 AB (E is the mid point of BC)
AD = 2AB (ABCD is a parallelogram therefore AD = BC)
$AB=\frac{1}{2}AD$