practice problems pending

 Q1) Let (ABCD) be a rectangle such that (BC = 3AB). (P) and (Q) are points on the side (BC) such that

BP = PQ = QC.
Show that

\angle DBC + \angle DPC = \angle DQC.

S1.



Q2.
In a quadrilateral ABCD, given that angle A + angle D = 90. Pr. th. AC^2 + BD^2 = AD^2 + BC^2.
S2.


BC^2 = OB^2 + OC^2
AD^2 = OA^2 + OD^2


BC^2 + AD^2 = OB^2 + OC^2 + OA^2 + OD^2
Look at RHS:
OA^2 + OC^2 = AC^2
OB^2 + OD^2 = BD^2
H.P.

Q3.

 In ( \triangle ABC ), (BM) and (CN) are perpendiculars from (B) and (C) respectively on a line passing through (A). If (L) is the midpoint of (BC), prove that

[
ML = NL.
]
S3.
Method 1:


Let LP be perpendicular to the same line passing through A.
LP || BM || CN
According to Intercept theorem, if 3 parallel lines cut 2 equal segments from a traversal, they will do the same with any other traversal.
So MP = PN
Now we will show triangle LMP congruent to LNP.
LP is common side.
Angle P is 90 in both.
MP = PN.
=> LM = LN.

Method 2: co-ordinate geometry

Let the line through A be x-axis and let A be (0,0).
B = (x1,y1) C = (x2,y2)
M = (x1,0) N = (x2,0)
L = x1+x2/2,y1+y2/2
Clearly LM = LN
H.P.

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