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 }
Then | ar(?DFE) |
ar(?CFB) |
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.
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