A nice geometry problem from New Zealand

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[pwmiles]

As Kuantum Alex points out, you can do it by Menelaus's theorem.
Take the triangle CXY and the line BPD.
Form ratios on the triangle sides, of the distances to the vertices from where they are cut by the line
I.e. BX/BC, PY/PX, DC/DY
Menelaus's theorem says the product of these is 1 i.e.
(BX/BC) (PY/PX) (DC/DY) = 1
But BX/DY=1, DC/BC=1leaving
PY/PX=1
Thus, P bisects XY or equivalently, the bisector of XY falls on the diagonal BD

[ethanyap]

Sine rule.
BX/sin angle BPX = YD/sin YPD (same angle and side length)
So PY/sin angle YDP = XP/sin angle XBP
So PY/sin 135 = PX/sin 45 and we're done

[crtwrght]

The similar triangle argument was nice. I did it with coordinates, setting D=(0, 0) and choosing the square to have sidelength 1 and BX=DY=x gives Y=(-x, 0) and X=(1, 1-x). The midpoint of XY is therefore ((-x+1)/2, (0+(1-x))/2)=0.5(1-x, 1-x) which is clearly on the line DB since it's the diagonal of a square and thus has gradient 1 starting from the origin.

[italyball]

You could have used the generalized ASA theorem to cut through the whole process of describing the two triangles as similar. For the rest, interesting video!

[TedHopp]

Once we have two congruent angle pairs, we can immediately conclude that the triangles are congruent by AAS. Then we can immediately conclude that sides XP and YP are congruent by the CPCT theorem. No need for that side trip through similar triangles.

[luggepytt]

First (like Michael did) notice that the length of Qx is equal to the length of Bx, because QBx is a 45°-45°-90° triangle. Hence it is also equal to yD.

Draw a horizontal line to the left from Q until it meets AD. Call this point S.
Draw a vertical line downwards from Q until it meets DC. Call this point T.

Now erase all parts of the square ABCD that are outside of the square DSQT.
Also erase the part QB of the blue diagonal.

The figure that is left - the square DSQT plus the two extensions Qx and Dy plus the blue diagonal DQ and the yellow line yx - has 180° rotational symmetry.
Therefore, P must be the midpoint of the diagonal DQ, and also the midpoint of yx.

[ffggddss]

Interesting problem, indeed! My first try would be to use analytic geometry.

Let C be the origin, with x-axis C-toward-B, y-axis C-toward-D. [To picture this in traditional position, rotate the diagram 90º clockwise.]
Let the side of the square be the unit of measure (ABCD is then a unit square), and call the choice of distance XB, a.

Then 0 < a ≤ 1, and the coordinates of X are (1–a, 0); of Y are (0, 1+a).
Diagonal BD lies along the line x + y = 1; XY lies along x/(1–a) + y/(1+a) = 1.

They intersect at
x/(1–a) + (1–x)/(1+a) = 1; (1+a)x + (1–a)(1–x) = 1 – a²
x = ½(1–a), y = 1 – x = ½(1+a)
which is the midpoint of XY, its coordinates being exactly halfway between those of X and Y.
QED

I bet there's a nicer, purely geometric proof that Michael will have for us...

And so he did. Nice!
Note: at ≈5min, you need only 2 equal pairs of corresponding ∠s to conclude similarity of two ∆s.

Fred

[geaninatudose]

YAX isosceles and YAX=90, then AYP=45 which is the same as ADC hence DYAP inscrptible so YDA=YPA=90 . hence AP altitude in isosceles, thus P midpoint of XY.

[wojteksocha]

You can just simply say that YDXQ is a parallelogram, so its diagonals intersect each in other in theirs mid point

[ddognine]

Using analytic geometry, assume that ABCD is a unit square with D at the origin. Also assume X is defined as the point (1, a). Thus one can easily solve for the equation of the line for XY then find the intersection, P, of XY and BD, which is the point (a/2, a/2). Since this is exactly one half of the the length of CX, it is simple to show by similar triangles that the lengths YP and PX are equal.

[wesleydeng]

Draw a line through Y and perpendicular to CD. Extend BD to intersect this line at point T. Easy to see YT=YD=BX. So, △YPT and △XPB are congruent. Therefore YP=XP.

[kuantumalex]

You can use for that problem Menalaus Theorem

[Bodyknock]

Am I missing something or doesn’t angle-side-angle imply the two triangles are congruent so you would immediately get the sides XP and YP are equal?

[Grizzly]

After showing that ∠QPX ≅ ∠DPY, it would be more economical to state that ∆QPX ≅ ∆DPY by the SAA postulate (XQ ≅ YD and ∠PQX ≅ ∠PDY as shown already), so XP = YP by CPCTC.

[s]

Wouldn't it be easier to argue that since there is no restriction on the lenght of BX or YD except that they are equal, by setting them both to 0 XY collapses to the diagonal and the midpoint would be on said diagonal? Also you can set x to C and YD = DC and the midpoint would be D also on the diagonal.

[galveston]

You can easily show this in Catesian coordinates. Write the equation line for BY and intesct. With the line y=x, find the coordinates of P and then XP=YP easily follows.

[yakovify]

Here is how I did it, start with x and y at the corners, then move x down and y left at the same velocity a. Their average position is exactly their midpoint and moves down with velocity a/2 and left with velocity a/2. Since the vertical and horizontal velocity are the same it's velocity is tangent to the diagonal line. Thus since the midpoint started on the line, the midpoint stays on the line always. Thus it is the intersection of the line y to x with the diagonal.

[emersonschmidt]

A straight forward aplication of sin law show it: since BX=DY, angle BPX = angle DPY and sin(45°)=sin(135°).

[highthrob]

You can create 2 small squares with diagonals YD and BX respectively. These squares have sides sqrt(1/2)*BX. This shows that if you slide YX along DB in direction of B by sqrt(1/2)*BX you get a situation that is point symmetrical around the center of the square. the translated P will be in this center, from which follows that YP = XP.

[cr]

There is another way that involves first showing triangle YAX is a 45-45-90 isosceles triangle, then showing AP is perpendicular to XY. Angle APX is 90 degree because angle AXP=ABP=45 degrees and therefore ABXP is co-cyclic which means ABX+APX=180 degrees.