Question

ABCD is a parallelogram the circle through A, B and C intersects cd at e prove that ae is equals to CD

Answer 1

Step-by-step explanation:

Given : ABCD is a parallelogram.  

To Prove : AE = AD.

∠AED + ∠ABC = 180° [Sum of opp. angles] ... (i)  

But ∠ACD = ∠ADC = ∠ABC + ∠ADE  

⇒ ∠ABC + ∠ADE = 180° [From (i)] ... (ii)  

From (i) and (iii)  

∠AED + ∠ABC = ∠ABC + ∠ADE

⇒ ∠AED = ∠ADE  

⇒ ∠AD = ∠AE [Sides opposite to equal angles are equal]

Hence proved

Answer 2

Given : ABCD is a parallelogram while ABCE is a cyclic quadrilateral.

To prove : AE = AD

Proof : Since ABCD is a parallelogram therefore,

∠ B = ∠ ADC [ Opposite angles are equal ] __(1)

Since ABCE is cyclic quadrilateral,

∠ E + ∠ B = 180° [ Sum of opp. ∠s ] __(2)

Substituting (1) in (2) we get,

∠ E + ∠ ADC = 180°

∠ E + (180° - ∠ ADE) = 180°      [∵ ∠ ADE + ∠ ADC = 180° ( Linear Pair )]

∠ AED = ∠ ADE

Hence, AD = AE ( Opp. sides of equal angles are equal )

Q.E.D