practice problems pending

1) ABCD is a parallelogram. Through C, a straight line RQ is drawn outside the figure and AP, BQ, DR are drawn ⟂ to RQ.

Show that:
DR + BQ = AP.
S1)

[RMO]
Key idea: DR || BQ so they can form a trapezium. Since LHS is DR + BQ which also appears in Trapezium midline theorem we will try to use Trapezium midline theorem which says this:
If AB || CD in trapezium ABCD, then line EF joining midpoints of AD and BC is || to AB,CD and also
EF = (AB + CD)/2.

Let O be intersection of diagonals AC and BD in the ||gram ABCD.
O is their midpoint.
Let OM be perpendicular to RQ.
DR || BQ and DBQR is a trapezium.

OM || DR || BQ and OM goes through midpoint of BD => M is the midpoint of RQ.
OM = (BQ + DR)/2

We are almost there now.
If we could show that OM = AP/2, we are done.

OM || AP
In Triangle APC, it's a strong hint to use Triangle midpoint theorem.
O is midpoint of AC and OM || AP => M is midpoint of PC.
=> OM = AP/2
H.P.

Another solution:

Drop a perpendicular DO from D to AP.
DRPO is a rectangle => DR = PO
AOD is congruent to BQC using AAS, AD = BC, Angle Q = Angle O, Angle A = Angle B
=> AO = BQ
AP = AO + OP = BQ + DR
H.P.

[IOQM]
Let's say it was a multiple choice question and you were asked DR + BQ = ? 
Then how to solve it quickly?

Since RQ can be any line through C, let's make it same as the line BC.
Now
BQ = 0
And distance from A to RQ = same as distance from D to RQ
=> AP = DR
DR + BQ = DR + 0 = AP


Q2

ABCD is a rectangle. Points M and N are on BD such that AM perpendicular to BD and CN perpendicular to BD. Prove that,

BM^2 - DM^2 = DN^2 - BN^2

S2: