The diagonals of a rectangle ABCD meet at 0. If ? BOC = 44°, then ? OAD is equal to :

From above given figure ,
Proceeding as in Q. No. 4, we can prove that AXCY is a parallelogram .
Similarly, BXDY is a parallelogram.
Now, AXCY is a parallelogram .
? AY || CX
[? Opposite sides of a parallelogram are parallel]
? PY || QX ?(1)
Also, BXDY is a parallelogram
? DX || BY [? Opposite sides of a parallelogram are parallel]
? PX || QY ?(2)
Thus, in a quadrilateral PXQY,
From (i) and (ii) ,
we have , PY || QX and PX || QY
? PXQY is a parallelogram.

We know that when two circle neither touch nor intersect and one lies outside the other, then 4 common tangents can be drawn.
For 4 common tangents , we get
? r₁ + r₂ < c₁c₂
Hence ,The number of tangents = 4

In a triangle ABC,
Given :- AB = 3 cm , AC = 5 cm and BC = 6 cm
We know that BD : DC = AB : AC
BD : DC = 3 : 5
? Divided BC = 6 in the ratio 3 : 5
BC = 3x + 5x = 6 ? 8x = 6 ? x = 6/8 = 0.75
? BD = 2.25, CD = 3.75.

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.