If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$.

Question

If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$.

My attempt to proof:

Suppose ($ACBD$) $= -1$ then $\frac{AC}{CB} = \frac{AD}{BD}$ then $AC \cdot BD = CB \cdot AD$. Also, since $O$ is a midpoint for $AB$ then $AO = OB$.

\begin{align*} AC \cdot BD &= CB \cdot AD \\ \Longrightarrow (AD + DC) \cdot BD &= (CD + DB) \cdot(AO + OD) \\ \Longrightarrow AD \cdot BD + BD \cdot DC &= CD \cdot AO + DB \cdot OD + DB \cdot AO + CD \cdot OD \\ \Longrightarrow AD \cdot BD - CD \cdot OD &= CD \cdot AO + DB \cdot DC + DB \cdot OD + DB \cdot AO \\ &= CD \cdot AO + DB \cdot DC + DB \cdot AD \\ &= CD \cdot AO + DB \cdot AC \\ &= (CA + AD) \cdot AO + (DA + AB) \cdot AC \\ &= CA \cdot AO + AD \cdot AO + DA \cdot AC + AB \cdot AC \\ &= AC \cdot (AB - AO) + DA \cdot (AC - AO) \\ &= AC(OB) + DA(OC) \end{align*} I'm stuck in this. Is my proof wrong? how to make the right hand side $0$?

Answer 1

Rephrasing of problem statement

Let $ A C = a, CB = b, BD = c$.
We are given that $ac = (a+b+c)b$.
WTS $(a+b+c)c = (b+c)(c+ (a+b)/2)$.

Taking the difference of the 2 sides, and multiplying by 2
WTS $ -ab + ac - b^2 - bc = 0$.
Notice that this is exactly the condition $ac = (a+b+c)b$. Hence we are done.

---

Note: If so desired, you can rephrase this in terms of $AC, CB, BD$, and use only those line segments, to make it clear what we should be focusing on.
Introducing line segments like $OA, OC$ etc makes it harder to manipulate, ad we should just replace them by $OA = (AC+CB)/2$, etc.

Answer 2

Sketch of another proof.

Construct a point $X$ such that $CX\perp AB$. Let $H$ be the orthocenter of the triangle $ABX$ and $XC$, $AM$, $BN$ are the corresponding altitudes. Then,

• $M$, $N$ and $D$ are collinear (hint: use Ceva's and Menelaus' theorems);

• the quadrilaterals $ANMB$ and $ONMC$ are cyclic (hint: nine-point circle).

Thus, $DA\cdot DB=DM\cdot DN=DO\cdot DC$, as desired.

Remark. Of course, this is much more complicated than direct computation via lengths of segments. However, I thibk it's, firstly, very nice proof (you can play with this constrcuction more) and, secondly, it illustrates the principle of passing to the higher diimensions which is quite useful (see proofs of Desargues' theorem, for example).