practice problems pending

Q1. The diagonals of a parallelogram (ABCD) intersect at (O). A line through (O) intersects (AB) at (X) and (DC) at (Y). Another line passing through (O) intersects (AD) at (P) and (BC) at (Q). Prove that (XQYP) is a parallelogram.

S1.

Triangles OAX and OCY are congruent.
Why?
OA = OC
Angle CYO = AXO
Angle AOX = YOC
=> OX = OY
=> O bisects XY

Similarly O bisects PQ.
So in quadrilateral XQYP, the diagonals bisect each other. Hence it's a ||gram. H.P.




So it gives us a property about ||gram that a line passing through diagonal midpoint and ending on opposite sides is also bisected at the diagonal midpoint.

Q2.

Prove that the feet of perpendiculars drawn from the vertices of a parallelogram onto its diagonals are the vertices of another parallelogram.

S2.


Let the feet of perpendiculars from D,B on AC be Q,P.
We will try to show that OP = OQ so that we can show that O bisects PQ.
Similarly, O will also bisect the other line joining the other 2 feet of perpendiculars.
And in that case diagonals will bisect each other and hence those feet will form a ||gram.

Ok, so trying to prove OP = OQ.
In Triangles OQD and OPB:
OD = OB
Angle OPB = OQD
Angle DOQ = BOP
So triangles OQD and OPB are congruent.
=> OP = OQ
H.P.


Q3. In an equilateral △ABC, a point P is taken in the interior of △ABC such that

(PA^2 = PB^2 + PC^2)

Find ∠BPC.



S3.
Since the problem hints at Pythagoras and has an equilateral triangle, let's try to rotate triangle BPC  around B by 60 deg.
Triangle BPC congruent to BP'A.
BC moves to BA after 60 deg rotation.
BP = BP' => BPP' is equilateral since B is already 60 deg.
=> PP' = BP = BP'.
PC = P'A.
So substitute these in the original equation:
PA^2 = PP'^2 + P'A^2
=> PA is hypotenuse and P' has the 90 deg angle => PP'A = 90 deg.
Angle BPC = Angle BP'A = BP'P + PP'A = 60 + 90 = 150 = Answer








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