Question

Question 6
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Answer 1

Given AE = AD.
To prove: AE = AD.
Construction: Draw a circle which passes through ABC and intersect CD (or CD produced) at E.
Proof: For fig (i)
$\angle AED+\angle ABC={180}^{\circ }$ [opposite angles of cyclic quadrilateral].....(i)
$\angle ABC=\angle ADC$ [opposite angles of parallelogram are equal].........(ii)
$\angle ADC+\angle ADE={180}^{\circ }$ ($\because$ Linear Pair) .......(iii)
Substituting (ii) in (iii) we get
$\angle ABC+\angle ADE={180}^{\circ }$ ........(iv)

From (i) and (iv)
$\angle AED+\angle ABC=\angle ABC+\angle ADE$
$\Rightarrow \angle AED=\angle ADE$
$\Rightarrow \angle AD=\angle AE$ [Sides opposite to equal angles are equal]
Similarly we can prove for Fig (ii) proved.