Given: ABCD is a parallelogram in which AC and BD are diagonals which intersect each other at O.
To Prove: AB² + BC² + CD² + DA² = AC² + BD²
Proof: Since diagonals of parallelogram bisect each other.
Therefore, BO and DO are medians of ∆ABC and ∆ADC, respectively.
We know that the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.
Now, in ∆ABC,
∵ OB is a median

We have,
AB ⊥ BC and DM ⊥ BC
⇒    AB || DM
Similarly, we have
CB ⊥ AB and DN ⊥ AB
⇒    CB || DN
Hence, quadrilateral BMDN is a rectangle.
∴    BM = ND
(i) In ∆BMD, we have
∠1 + ∠BMD + ∠2 = 180°
⇒ ∠1 + 90° + ∠2 = 180°
⇒    ∠1 + ∠2 = 90°
Similarly, in ∆DMC, we have
∠3 + ∠4 = 90°
Since BD ⊥ AC. Therefore
∠2 + ∠3 = 90°
Now, ∠1 + ∠2 = 90°and ∠2 + ∠3 = 90°
⇒    ∠1 + ∠2 = ∠2 + ∠3
⇒    ∠1 = ∠3
Also, ∠3 + ∠4 = 90° and ∠2 + ∠3 = 90°
⇒    ∠3 + ∠4 = ∠2 + ∠3 ⇒ ∠2 = ∠4
Thus, in ∆'s BMD and DMC, we have
∠1 = ∠3 and ∠2 = ∠4
Therefore, by using AA similar condition
                  



                  

               
 
(ii) Proceeding as in (i), we can prove that
                  

Given: An acute triangle, acute angled at B and AD ⊥ BC.
To Prove : AC² = AB² + BC² — 2BC.BD
Proof: In ∆ABD, we have
AB² = AD² + BD² ...(i)
[Using Pythagoras theorem]
In ∆ADC, we have
AC² = AD² + DC²
⇒ AC² = AD² + (BC - BD)²
⇒ AC² = AD² + BC² - 2BC.BD + BD²
⇒ AC² = (AD² + BD²) + BC² - 2BC.BD
⇒ AC² = AB² + BC² - 2BC.BD
[from (i), we have AD² + BD² = AB²]

Given: An obtuse triangle ABC, obtuse angled at B, and AD ⊥ CB produced.
To Prove : AC² = AB² + BC² + 2BC.BD
Proof: In right triangle ADB, we have
AB² = AD² + BD² ...(i)
[Using Pythagoras theorem]
In ∆ADC, we have
AC² = AD² + DC²
⇒ AC² = AD² + (DB + BC)²
⇒ AC² = AD² + DB² + BC² + 2DB.BC
⇒ AC² = AB² + BC² + 2BC.BD
[from (i), we have AB² = AD² + BD²]