analytic geometrygeometry
Quadrilateral ABCD is inscribed in circle O.
AB and DC intersects at point E, AD and BC intersect at point F.
The area of traingle ABC ,traingle ADC are equal, prove that EH = HF.
I wonder whether this problem can be solved by pencil of conics that pass A,B,C,D.
Which I suppose EF is the X-axis, then EH , FH can be the two root of the formula.
I don't know how to take advantage of the given condition that the area of triangles are equal.
Any help is appreciated! Thanks:)
From the diagram, $AB=CD=(r_1+r_2)\cos\alpha$.
$AC=2r_1\cos\alpha$ and $BD=2r_2\cos\alpha$. Therefore, $AB+CD=AC+BD$, and $ABDC$ is a tangential quadrilateral.
$\blacksquare$
Below is a "full" solution to the problem. A hint, if you don't feel quite ready to see the solution yet, is to consider the Miquel point of the complete quadrilateral $ABCD$.
Let points $A,B,C,D,E,F,M,P,Q,X,Y$ be defined as in the question. Define $\gamma$ to be the circumcircle of $ABCD$, and redefine $O$ to be its centre. Let the circle with centre in $Y$ through $X$ be called $\omega$.
Lemma 1 (Miquel Point of Cyclic Quadrilaterals): Let $Z$ be the intersection of lines $OX$ and $EF$. Then $Z$ is the miquel point of the complete quadrilateral $ABCD$. In particular, $Z$ is the image of $X$ under inversion with respect to $\gamma$.
Lemma 2 ($EFX$ is self-polar with respect to $\gamma$): $X$ and $Z$ lie on the normal line from $O$ to $EF$.
These two facts turn out to be sufficient in explaining the tangency in the question as follows:
Due to lemma 2 $\angle MZX=\angle MZO = \pi/2$. Therefore, since $MX$ is a diameter in $\omega$, $Z$ lies on $\omega$ by the converse of Thale's theorem.
But now lemma 1 tells us that $X$ and $Z$ are inverse images under inversion in $\gamma$, implying that $|OX||OZ|=r^2$, where $r$ is the radius of $\gamma$. Hence the power of $O$ with respect to $\omega$ is $r^2$. Now suppose a tangent to $\omega$ through $O$ intersect $\omega$ at $T$. Then by power of a point $|OT|^2=r^2$, so $T\in\gamma$. But $T\in\omega$ by assumption, so $T=P$ or $T=Q$.
Said in another way: the tangents from $O$ to $\omega$ are exactly the lines $OP$ and $OQ$.
Now you may realise that this is rather similar to what we want to prove, that is, the tangents from $Y$ to $\gamma$ are the lines $YP$ and $YQ$. It turns out that these two statements are actually equivalent. You may try to prove it yourself. The configuration is called orthogonal circles. At any rate, this explains the problem posed in the original post.
All of the concepts/lemmas that I have used are defined/proven thoroughly in chapters 8 to 10 of Evan Chen's Euclidean Geometry in Mathematical Olympiads . It is a good introduction to many of the more advanced techniques used in olympiad geometry, and I wholeheartedly recommend it if you are preparing for maths olympiads.
Best Answer
If we denote the intersection point of $BD$ and $AC$ by $L$, then $L$ will be the midpoint of $BD$.
Then, $\frac{BC}{AD}=\frac{CL}{LD}=\frac{CL}{BL}=\frac{CD}{AB}$
$\triangle EBC\sim \triangle EDA$
$\Rightarrow \frac{CE}{AE}=\frac{BC}{AD}$
$\Rightarrow \frac{CE}{AE}=\frac{CD}{AB}$
$\Rightarrow \frac{CE}{CD}=\frac{AE}{AB}$
Applying Menelaus' Theorem on $\triangle AED$ considering $BCF$ as the transversal gives,
$\frac{AB}{BE}\cdot \frac{EC}{CD}\cdot \frac{DF}{AF}=1$
$\Rightarrow \frac{AB}{BE}\cdot \frac{AE}{AB}\cdot \frac{DF}{AF}=1$
$\Rightarrow \frac{AE}{BE}=\frac{AF}{DF}$
Hence, $BD\parallel EF$.
Since $AC$ bisects $BD$, $AC$ must bisect $EF$ as well. Thus, $H$ is the midpoint of $EF$.