homework pending

 Q1. In △ABC, AD, BE and CF are concurrent lines. P, Q, R are points on EF, FD, DE such that DP, EQ and FR are concurrent. Prove that AP, BQ and CR are also concurrent.
S1.
Prerequisite: Standard and trigonometric form of Ceva's theorem.

Standard form gives us:
AF/FB * BD/DC * CE/EA = 1
and
EP/PF * FQ/QD * DR/RE = 1

Using the trigonometric form, we need to prove that:
Sin(CAP)/Sin(PAB) * Sin(ABQ)/Sin(QBC) * Sin(BCR)/Sin(RCA) = 1

Look at triangles EAP and PAF and compute their area ratio:
[EAP]/[PAF] = AE * AP * Sin(CAP)/AF*AP*Sin(PAB) = EP/PF
Similarly do for others.

Multiply both sides of all 3.
You will get your desired relation.
H.P.

Q2.

If (X) and (Y) are variable points on the sides (CA, AB) of (\triangle ABC) such that
CX/XA + AB/AY = 1

prove that (XY) passes through a fixed point.

S2.
CX/XA > 0 since everything is positive.
=> AB/AY < 1
=> AY > AB
=> Y lies on AB extended towards B.
=> AB = AY - BY

=> AB/AY = 1 - BY/AY

From here: we have 2 cases for X.

Case 1: X lies internally on AC.

CX/XA + AB/AY = 1
=> CX/XA + 1 - BY/AY = 1
=> CX/XA = BY/AY__________[1]


Now, let XY intersect BC at some point Z internally.
Why externally?
Since X is internal, Y is external, Z has to be internal.
Why?
Line XYZ, if it has entered the triangle, it has to exit as well.
So there will be 2 internal points which are X,Z.

From Menelaus:
AX/XC * CZ/ZB * BY/YA = 1

Use [1] to get:
CZ/ZB = 1 => CZ = ZB
=> Z is the midpoint of BC
So the fixed point here is the midpoint of BC.

Case 2:
X lies externally BC.

CX/XA + AB/AY = 1
Earlier we showed that Y lies externally on AB towards B.
From the same logic:
X will lie externally on AC towards C.
Why?
Then CX/XA < 1 Else it would be more than 1 if X lies towards A.

AC = AX - CX
=> AC/AX = 1 - CX/AX
=> CX/AX = 1 - AC/AX
=> 1- AC/AX + AB/AY = 1
=> AC/AX = AB/AY

So Triangles ABC and AYX are similar from SAS criteria.
=> CB and XY are parallel.
=> They will never meet.

This is an edge case we need to consider.

Q3.

A straight line cuts the sides 

AB,BC,CD,DA of a quadrilateral in P,Q,R,S. Prove that AP/PB⋅BQ/QC⋅CR/RD⋅DS/SA=1

S3.
This hints at generalized Menelaus.
For any Convex Polygon we can apply generalized Menelaus.
Proof is the same.
Drop perpendiculars on the traversal from all vertices.
Case 1:
Let A,B be 2 vertices which are being cut internally by the traversal at P.
Let foot of perpendicular from A to traversal be Ha and from B it's Hb.
Triangles AHaP and BHbP are similar.
=> AP/PB = AHa/BHb = height from A/height from B
Keep doing it for all vertex pairs and multiply them all and eventually everything cancels out to give you 1.

Side notes:
A traversal can cut a convex polygon either 0 or 2 times internally. All other cuts will be external.
Why?

Definition of Convex polygon:
If 2 points lie inside the shape then the line joining them lies completely inside the shape.
Let a traversal enter the polygon at point A and exit at point B.
Now if enters the polygon again at C then there is a problem.
Pick a point between A and B and join it with the point just next to C.
That line doesn't lie completely inside the shape.
H.P.

So when doing Menelaus, remember that a traversal will either cut the convex polygon twice or never at all.

Q4.

In quadrilateral ABCD, let AB and CD meet at E, AD and BC meet at F. Then prove that the midpoints of AC,BD and EF are collinear.

S4.

This problem is about Newton-Gauss line in a complete quadrilateral.
We need to show that midpoints of diagonals of a complete quadrilateral are always colinear.

But what is a complete quadrilateral?
It is:
4 lines intersecting each other at 6 points such that no 3 lines pass through one point.

Here those 4 lines are:
AB, CD, AD, BC
The 6 intersection points are A,B,C,D,E,F.

A complete quadrilateral has 3 diagonals.
Here they are:
AC,BD,EF
Basically pick one vertex each from opposite sides.

Now coming to the main task, proving that midpoints of AC,BD,EF are colinear.
Let midpoints of sides of triangle EBC be X,Y,Z