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 .

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 figure , we have
Since every exterior angle is equal to sum of interior opposite angles,
So, ? a = A + B, ? b = B + C and ? c = A + C
We know that , A + B + C = 180°
? ?a +?b + ?c = 2(A + B + C) = 2 × 180° = 360°.

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.