In the given figure DE||BC and AD : DB = 5 : 4,

Then	ar(?DFE)
	ar(?CFB)

In ||gm ABCD , we have
Since diagonals of parallelogram(||gm ) bisect each other,
? M will be the mid-point of each of the diagonal AC and BD
? In ?ABC AB² + BC² = 2(AM² + MB²) ......... ( 1 ) [Appolonius Theorem]
In ?ADC AD² + CD² = 2(AM² + DM²) ......... ( 2 )
= 2(AM² + MB²) [? DM = BM]
Adding equations ( 1 ) and ( 2 )
AB² + BC² + CD² + DA² = 2(AM² + MB² + 2(AM² + MB² = 4AM² + 4MB²
= (2AM)² + (2MB)²= AC²+ BD². { ? AM = MC , MB = MD }

We know that In center of every triangle ( acute , obtuse and right angle ) lies in its interior.
Hence required answer will be any triangle .

From above figure , we can see that
Join B and D and produce BD to E
Then, p + q = ? and s + t = x
We know that , Exterior angle = Sum of two interior angle
Now, s = p + ? (Exterior angle of a ? prop.)
Similarly, t = q + ? (Exterior angle of a ? prop.)
Adding, s + t = p + q + ? + ?
Putting s + t = x and p + q = ? ,
Hence x = ? + ? + ?

From the given figure, In ?ABC
We know that , Exterior angle = Sum of two interior angles
?1 = ?A + ?5 ................. ( 1 )
and ?2 = ?A + ?6 ................. ( 2 )
Adding equations ( 1 ) and ( 2 ).
?1 + ?2 = 2?A + ?5 + ?6
= 2?A + (180° ? ?A) = ?A + 180°
The given question can be restated as the sum of two exterior angles exceeds ?A of the ?ABC by 2 right angles.

From given figure , we have
In quadrilateral ABCD , we can see that
Diagonal AB = Diagonal CD ......... ( ? )
AO = OB ......... ( ? )
DO = CO ......... ( ? )
?DOB = 90° ......... ( ? )
From ( ? ) , ( ? ) , ( ? ) and ( ? ) , we get
From above , it is clear that In quadrilateral ABCD , diagonals are equal and they bisect each other at 90° .
? Quadrilateral ABCD is square.

From given figure , we have
Let, (x + 1) be the hypotenuse and perpendicular = x - 1 , base = x.
By pythagoras theorem ,
? (x + 1)² = x² + (x + 1)² ? x = 4
? Hypotenuse = 4 + 1 = 5