Given $\triangle BPD$ with $O$ the midpoint of $\overline{BD}$, project $P$ to $P'$ and show that $|OB||OP'|$ is independent of $|BD|$.

Question

This recent question (put on-hold for lack of context) presented the following:

Original Question. Let $ABCD$ be a square and let $P$ be a point inside it such that $|PD|=29$ and $|PB|=23$. Find the area of $\triangle APC$.

As @JeanMarie's answer indicates, the answer is (perhaps surprisingly) independent of the size of the square. In general, the prescribed area can be given as

$$|\triangle APC| = \frac14\,\left|\,|PB|^2-|PD|^2\,\right| \tag{1}$$

My approach (which I couldn't post because the question was closed) was as follows: Project $P$ to $P'$ on $\overline{BD}$; then $\triangle APC$ and $\triangle AP'C$ (which share base $\overline{AC}$ and have congruent corresponding heights) have equal areas, so we just need to find the latter.

Of course, that area is (writing $O$ for the center of the square): $$|\triangle AP'C| = \tfrac12 |AC||OP'| = |OA||OP'|=|OB||OP'| \tag{2}$$ With the far-right expression, we find that we can ditch the square context and the goal of computing an area, reformulating the question something like this:

Reformulated Question. For $\triangle BPD$ with $O$ the midpoint of $\overline{BD}$, let $P'$ be the projection of $P$ onto $\overline{BD}$. Show that $|OB||OP'|$ is independent of $|BD|$ (or, if you prefer, $\angle P$).

Specifically, taking the target to be a signed product (negative, if $\overrightarrow{OB}$ and $\overrightarrow{OP'}$ are oppositely-directed; positive otherwise) show that the value is given by $$|OB||OP'| = \frac14\left(\;|PD|^2 - |PB|^2\;\right) \tag{3}$$

My answer to the Reformulated Question appears below. Other approaches are welcome.

Answer 1

Apply the Pythagorean theorem to the right triangles BPP' and DPP' that share the height PP’,

$$PB^2 - (OB+OP')^2 = PD^2 - (OB-OP')^2 $$

which yields the result,

$$OB\cdot OP' = \frac14\left(PB^2 - PD^2\right)$$

Answer 2

A recall : The product of "algebraic lengthes" of a vector $\vec{u}$ with the projection of another vector $\vec{v}$ on the axis defined by vector $\vec{u}$ is their dot product $\vec{u}.\vec{v}$.

Therefore, it suffices to prove, with vector notations, that

$$\vec{OP}.\vec{OB}=\tfrac14\left(\vec{PD}^2-\vec{PB}^2\right)\tag{1}$$

In fact, (1) boils down to the well-known relationship (valid in any inner product space) :

$$\vec{u}.\vec{v} = \dfrac14\left(\|\vec{u}+\vec{v}\|^2 - \|\vec{u}-\vec{v}\|^2\right),$$

with $\vec{u}:=\vec{OP}$ and $\vec{v}:=\vec{OB}$.

(Indeed $\vec{OP}+\vec{OB}=\vec{DP}$ and $\vec{OP}-\vec{OB}=\vec{BP}$.)

Answer 3

Define $b := \tfrac12|PB|$ and $d :=\tfrac12|PD|$, and let $M$ be the midpoint of $\overline{PB}$. Note that $\overline{OM}$ is a midpoint segment of $\triangle BPD$, so it has length $d$.

Now, since $\angle PP'B$ is a right angle, Thales' Theorem tells us that $P'$ lies on the circle with diameter $\overline{PB}$ (with center $M$).

Finally, we simply observe that the signed product $|OB||OP'|$ gives the power of point $O$ with respect to $\bigcirc PP'B$, so that

$$|OB||OP'| = |MO|^2 - (\text{radius})^2 = d^2 - b^2 = \frac14\left(\,|PD|^2-|PB|^2\,\right) \tag{$\star$}$$

as desired. $\square$

Answer 4

Circumscribe a circle around triangle $BP'P$, which will have a centre $S$ in the middle of $PB$. Let $r=|PB|/2$ be the radius of that circle and let $OS$ intersect the circle in points $X$ and $Y$:

Note that $|OS|=|PD|/2$ as a midsegment. If we for a moment assume that $|PB|\ge|PD|$, then $O$ is inside the circle.

Now, it is well known that:

$$\begin{array}{rcl}|OP'|\cdot|OB|&=&|OX|\cdot |OY|\\&=&(r-|OS|)(r+|OS|)\\&=&r^2-|OS|^2\\&=&(|PB|/2)^2-(|PD|/2)^2\\&=&\frac{1}{4}(|PB|^2-|PD|^2)\end{array}$$

The claim then follows because the product "as vectors", for $O$ within the circle, is taken with the negative sign.

The consideration is very similar in the case $|PB|\le|PD|$, except that, in that case, $O$ is outside the circle.