euclidean-geometrysimilitudetriangles
This question was asked in a scholarship test for class 10th.
In the given figure, ABCD is a trapezium, AD is parallel to BC and diagonal BD is parallel to XY. 3AD=BC. Find the ratio of AY to AX.
My attempt (By using Similarity of Triangles):-
YD/DC=YA/AX
YD/DC=BX/BC
YA/AX=BX/BC
Let AD=y, BC=3y.
On producing AB and CY, let them intersect at point T.
TA/AB=TD/DC=AD/BC=1/3
So, AB=2TA and CD=2TD
When I'm too lazy to find a suitable chain of established theorems, I tend to resort to algebraic computations. I like using homogeneous coordinates and projective geometry for this. I dislike square roots, as you'd get them when computing the incircle from the circumscribed triangle. So I'll start with that incircle. Without loss of generality, the incircle of $\triangle BCD$ is the unit circle, with $E=[0:1:1]$ at the top to match the figure. Then the other two points of contact can be chosen using the tangent half-angle formulas in such a way that they can represent any point on the unit circle except $E$.
$$E=\begin{bmatrix}0\\1\\1\end{bmatrix}\qquad T=\begin{bmatrix}2t\\t^2-1\\t^2+1\end{bmatrix}\qquad U=\begin{bmatrix}2u\\u^2-1\\u^2+1\end{bmatrix}$$
The tangent line at each of these points can be obtained by multiplying the matrix of the unit circle with the vector. Since that matrix is $\operatorname{diag}(1,1,-1)$ (representing $x^2+y^2-1=0$) you simply negate the last coordinate. Then the points where the tangents intersect can be computed as cross products between the lines.
\begin{align*} B&\sim\begin{bmatrix}2t\\t^2-1\\-t^2-1\end{bmatrix}\times\begin{bmatrix}2u\\u^2-1\\-u^2-1\end{bmatrix}=2(t-u)\begin{bmatrix}t+u\\tu-1\\tu+1\end{bmatrix}\\ C&\sim\begin{bmatrix}2t\\t^2-1\\-t^2-1\end{bmatrix}\times\begin{bmatrix}0\\1\\-1\end{bmatrix}=2\begin{bmatrix}1\\t\\t\end{bmatrix}\\ D&\sim\begin{bmatrix}2u\\u^2-1\\-u^2-1\end{bmatrix}\times\begin{bmatrix}0\\1\\-1\end{bmatrix}=2\begin{bmatrix}1\\u\\u\end{bmatrix} \end{align*}
Assuming $T\neq U$ and therefore $t\neq u$, we can cancel all these coefficients in front of the vectors and continue calculations with the simple representatives. Next we need a reflection in the symmetry axis of the trapezium. You could compute this as the uniquely defined projective transformation which exchanges $C$ with $D$ and exchanges the ideal circle points $I=[1:i:0]$ and $J=[1:-i:0]$ (which is the mark of any orientation-reversing similarity). But I'd rather avoid the complex numbers here, and instead go with the inhomogeneous transformation
$$\begin{pmatrix}P_x\\P_y\end{pmatrix}\mapsto \begin{pmatrix}C_x+D_x-P_x\\P_y\end{pmatrix}= \begin{pmatrix}\frac1t+\frac1u-P_x\\P_y\end{pmatrix}$$
Written in homogeneous coordinates and avoiding divisions by multiplying everything with $tu$ this becomes $P\mapsto RP$ with the reflection matrix
$$R=\begin{bmatrix}-tu&0&t+u\\0&tu&0\\0&0&tu\end{bmatrix}$$
We can use this to find $A$ as
$$A=RB=\begin{bmatrix}t+u\\tu(tu-1)\\tu(tu+1)\end{bmatrix}$$
To compute the angular bisector through $A$ would again entail a square root, but we can also define that same line as the line through $A$ and the reflection of the origin $O=[0:0:1]$ (which is the center of the incircle).
$$l_{AF}\sim A\times(RO)=tu\begin{bmatrix}tu(tu-1)\\tu(t+u)\\(t+u)(1-tu)\end{bmatrix}$$
This line we intersect with $[1:0:0]$, the vector for the line $1x+0y+0=0$, to obtain $F$.
$$F\sim\begin{bmatrix}tu(tu-1)\\tu(t+u)\\(t+u)(1-tu)\end{bmatrix}\times\begin{bmatrix}1\\0\\0\end{bmatrix}=-(t+u)\begin{bmatrix}0\\tu-1\\tu\end{bmatrix}$$
Omitting this factor $t+u$ here assumes $t\neq-u$ or in other words $T,U$ are not mirror images with respect to the vertical axis through the unit circle. If they were, then $A$ and $B$ would be the same point, so I think this is again a fair assumption.
Now we need the circle through $A,C,F$, which can be computed as the conic through $A,C,F,I,J$ with $I=[1:i:0]$ and $J=[1:-i:0]$ as the ideal circle points. First define two degenerate conics through $A,C,I,J$, each given as the symmetric product of a pair of (complex) lines. Then the circle is the linear combination which also passes through $F$.
$$ Q_1=(I\times A)\odot(J\times C)=\tfrac12\bigl((I\times A)(J\times C)^T+(J\times C)(I\times A)^T\bigr) \\ Q_2=(J\times A)\odot(I\times C)=\tfrac12\bigl((J\times A)(I\times C)^T+(I\times C)(J\times A)^T\bigr) \\ Q\sim(F^TQ_2F)Q_1-(F^TQ_1F)Q_2=i u t^3 (u^2 + 1)\cdot\\ \begin{bmatrix} 2 u t^2 (tu + 1) & 0 & -t^2 (u^2 + 1) \\ 0 & 2 u t^2 (tu + 1) & -t (2t^2u^2 + u^2 - 1) \\ -t^2 (u^2 + 1) & -t (2t^2u^2 + u^2 - 1) & 2 u (tu - 1) (t^2 + 1) \end{bmatrix}$$
I have to concede that this circle is quite a beast. We have non-degeneracy conditions $t\neq 0$ and $u\neq 0$, since either of these would lead to one tangent being parallel to $CD$ and this to points at infinity. The point $G$ can be computed as
$$G\sim (D^TQD)C - 2(C^TQD)D=2(t-u)^2 \begin{bmatrix}t-u\\tu(tu+1)\\tu(tu+1)\end{bmatrix}$$
Now we need to express the isosceles condition. One possibility: $F$ lies on the orthogonal bisector of $AG$. To describe that bisector you can join the midpoint $M$ between $A$ and $G$ to the point $N$ at infinity orthogonal to $AG$.
$$M\sim G_3A+A_3G=2t^2u(tu+1)\begin{bmatrix}1\\tu^2\\u(tu+1)\end{bmatrix}\\ N\sim\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}(A\times G)= -2tu^2(tu+1)\begin{bmatrix}t\\1\\0\end{bmatrix}$$
Now three points are collinear if their determinant is zero.
$$\det(F,M,N)=\begin{vmatrix} 0 & 1 & t \\ tu-1 & tu^2 & 1 \\ tu & u(tu+1) & 0 \end{vmatrix} = 0$$
So $F$ does indeed lie on the perpendicular bisector, as claimed. Q.e.d.
You are right ! The problem does indeed involve Thales' theorem , and similarity:-
Note that we have:-
$\triangle OEC \sim \triangle DBC \implies \frac{EC}{BC}=\frac{OC}{DC} \tag{1}$ $\triangle OFB \sim \triangle ACB \implies \frac{BF}{BC}=\frac{OB}{AB} \tag{2}$
But by Thales' Theorem , the RHS of $(1)$ and $(2)$ are equal !
Therefore , $EC$ equals $BF$ , which implies $EB$ equals $CF$
Best Answer
It's unnecessary to extend $AB$ and $CY$ until they intersect at a point $T$. Instead, here's your diagram with an extra line of $BD$, and a few lengths, added to it:
In particular, setting $|AX|=e$ and $\frac{|AY|}{|AX|}=r$, then $|AY|=re$. With $AD \parallel BC$, we get $\triangle YAD \sim \triangle YXC$. Thus, with $|DC|=g$, we have $|YD|=rg$.
Setting $|AD|=f$ (instead of your $y$, to help avoid any possible confusion with $Y$), then $|BC| = 3|AD| = 3f$. Next, as you've already noted, with $BD \parallel XY$, then $\triangle CBD \sim \triangle CXY$, so $|BX|=r|BC|=3rf$.
Using $\triangle YAD \sim \triangle YXC$ again, then
$$\begin{equation}\begin{aligned} \frac{|YA|}{|AD|} & = \frac{|YX|}{|XC|} \\ \frac{re}{f} & = \frac{(r+1)e}{3(r+1)f} \\ \frac{re}{f} & = \frac{e}{3f} \\ r & = \frac{1}{3} \end{aligned}\end{equation}$$
Thus, $|AY|:|AX|=1:3$.
Alternatively, as explained in Joffan's answer, $ADBX$ is a parallelogram, so $|AD|=|XB| \; \to \; f = 3rf \; \to \; r = \frac{1}{3}$.