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I have the following question with me:

Let $ABCD$ be an isosceles trapezium with $AB$ parallel to $CD$. The inscribed circle of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the internal bisector of angle $DAC$ such that $EF$ is perpendicular to $CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that triangle $AFG$ is isosceles.

Figure

My try:

Define H to be the point of intersection of $CD$ and $AF$. I deduced very few points by simple properties of triangles

$$\frac{AH}{HF} = \frac{AC}{CF} $$ $$\frac{DH}{HC} = \frac{AD}{AC}=\frac{BC}{BD}$$

I also tried ptolemy and angle chasing but couldn't get anything. Any help please.

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  • $\begingroup$ It would be helpful if you add a diagram showing the points and shapes to explain this better $\endgroup$ – Krishna Sep 30 '18 at 7:03
  • $\begingroup$ What is this $M$ in "[l]et $ABCD$ be an isosceles trapezium with $AM$ parallel to $CD$"? $\endgroup$ – user10354138 Sep 30 '18 at 7:13
  • $\begingroup$ can anybody help me with drawing a geometry diagram on a computer and sorry its not M it is B, i have edited the question accordingly $\endgroup$ – saisanjeev Sep 30 '18 at 7:52
  • $\begingroup$ You can use geogebra @saisanjeev $\endgroup$ – ArsenBerk Sep 30 '18 at 7:59
  • $\begingroup$ @ArsenBerk I am finding it very difficult to construct the figure there $\endgroup$ – saisanjeev Sep 30 '18 at 8:07
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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 $E$.

$$E=\begin{bmatrix}0\\1\\1\end{bmatrix}\qquad T=\begin{bmatrix}2t\\t^2-1\\t^2+1\end{bmatrix}\qquad U=\begin{bmatrix}2u\\u^2-1\\u^2+1\end{bmatrix}$$

Figure

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.

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  • $\begingroup$ Sorry for my ignorance but what have you represented in those matrices? $\endgroup$ – saisanjeev Oct 1 '18 at 15:22
  • $\begingroup$ @saisanjeev: which ones? The column vectors are homogeneous coordinate vectors of points: $[x:y:1]$ or some multiple thereof. Except those that are lines $[a:b:c]$ representing $ax+by+c=0$. $\begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}$ represents the conic section $x^2+y^2=1$ i.e. the unit circle. $Q$ is the thircle through $A,C,F$. $R$ is a reflection in the symmetry axis of the trapezium. And the determinant in the end, with the vertical bars, has three points as columns to check for collinearity. This answer builds on a background in projective geometry, don't have a better idea yet. $\endgroup$ – MvG Oct 1 '18 at 17:17
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Starting with the posted figure, extend $AD$ to $K$, join $FK$, $FC$, draw $FD$ through to $M$, join $GK$, $GM$, extend $FE$ to $L$ on the incircle, and through $L$ draw a line parallel to $CG$.

Then granting, as seems to be the case, that $L$ lies on circle $ACF$, and that $M$ lies on the parallel through $L$, it would follow that$$\triangle FGM\sim\triangle FEC$$because$$\angle GFM=\angle EFC$$since they stand on equal arcs $GM$ and $CL$, and$$\angle FMG=\angle FCE$$ since they stand on common arc $GF$. triangle GFA is isosceles Therefore $\angle MGF=\angle CEF$ is right and $FM$ is a diameter of circle $ACF$.

And since $D$ lies on this diameter, it follows that$$DK=DC$$and$$DG=DA$$whence by SAS$$\triangle GDK\cong\triangle ADC$$

And finally, since$$\angle KGF=\angle FGC=\angle KAF=\angle FAC$$making$$KF=FC$$then by AAS$$\triangle KGF\cong\triangle CAF$$whence$$GF=AF$$and triangle $GFA$ is isosceles.

I am unable to prove that $L$ lies on circle $ACF$, and that $M$ lies on the parallel through $L$, but of several figures and attempts, each missing a link somewhere, this is the nearest I get to a purely geometric understanding of this interesting phenomenon.

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