If the sides of a triangle are produced then the sum of the exterior angles i.e., ? a + ? b + ? c is equal to :

As per given figure , we can see that
Since the diagonals of a rectangle are equal and bisect each other. Let AC and BD intersect at M. Therefore M is the mid-point of AC and BD and AM = DM
From ?AOC, OA² + OC² = 2(AM² + MO²) [Appollonius Theorem.] .......... ( 1 )
also in ?ODB, OB² + OD² = 2(MO² + DM²) = 2(MO² + AM²) .......... ( 2 )
From equations ( 1 ) and ( 2 ) .
? OA² + OC² = OB² + OD².

From above given triangle , It is clear that the altitudes AL, BM and CN of ?ABC intersect at H. Then H is the orthocentre of ?ABC.
In ?ABC, HL ? BC and BN ? CH.
Thus, the two altitudes HL and BN of ?HBC, intersects at A.
Hence the orthocentre of ?HBC is M .

From above given figure , we can see that
CE || BA and AC is the transversal.
? ?4 = ?1 (alternate interior ?s) .......... ( 1 )
Again, CE || BA and BD is the transversal.
? ?5 = ?2 (corresponding ?s) ................( 2 )
Adding equations ( 1 ) and ( 2 )
? ?4 + ?5 = ?1 + ?2
??ACD = ?A + ?B.

Let ABCD is a rhombus in which AB = BC = CD = DA = diagonal CA.
In triangle ABC ,
AB = BC = CA , therefore ?B = 60°
?D = ?B = 60°
?A=180 - ?B = 180° - 60° =120°
?C = ?A=120°
Hence ,The ratio of diagonals = ?2 : 1