# solution-verification | Prove that the lines $BP,A'N,C'M$ are concurrent

The problem

Let $$ABCDA'B'C'D'$$ be a rectangular parallelepiped. We denote the projections of the vertex $$B'$$ on the diagonals $$A'B, BC', A'C'$$ by $$M, N, P$$.

$$a)$$ Show that $$B'M$$ and $$B'N$$ are the distances from point $$B'$$ to the planes $$(A'BD')$$ and $$(ABC')$$ respectively

$$b)$$ Prove that the lines $$BP,A'N , C'M$$ are concurrent

My idea

Drawing

For point $$a)$$ we know that because $$ABCDA'B'C'D'$$ is a rectangular parallelepiped $$A'D' \perp (ABB'A')=> A'D' \perp MB'$$ and we also know $$MB' \perp A'B$$ so we demonstrated that $$MB' \perp (D'A'B)$$ which means that the distance from $$B'$$ to $$(A'BD')$$ is $$MB'$$

The same thing we do for the other distance:

we know that because $$ABCDA'B'C'D'$$ is a rectangular parallelepiped $$AB \perp (BCC'B')=> AB \perp B'N$$ and we also know $$B'N \perp BC'$$ so we demonstrated that $$B'N \perp (ABC')$$ which means that the distance from $$B'$$ to $$(ABC')$$ is $$B'N$$

Now for point $$b)$$ I thought of using the reciprocal of the theorem of Ceva so I calculated the ratios determined by $$M, N, P$$ in triangle $$BA'C'$$ using the theorem of the cathetus.

$$C'P=\frac{B'C'^2}{A'C'}, PA'=\frac{B'A'^2}{A'C'}$$ so $$\frac{PA'}{C'P}=\frac{B'A'^2}{B'C'^2}$$

Analogous, we get that $$\frac{PA'}{C'P}*\frac{C'N}{NB} \frac{MB}{A'M}= \frac{B'A'^2}{B'C'^2} \frac{B'C'^2}{B'B^2} \frac{B'B^2}{B'A'^2}=1$$

so we proved that the lines $$BP, A'N, C'M$$ are concurrent

I'm unsure if I used the reciprocal of the theorem of ceva right or if I missed any info I should know before using it.

I hope one of you can help me! Thank you!

Your solution looks correct to me.

$$b)$$

There is another solution which proves that $$C'M\perp A'B, A'N\perp BC'$$ and $$BP\perp A'C'$$.

Since $$\triangle{BB'C'},\triangle{C'B'M}, \triangle{BB'M}$$ are right triangles, we have $$BC'^2=B'C'^2+BB'^2\tag1$$ $$C'M^2=B'C'^2+MB'^2\tag2$$ $$MB'^2=BB'^2-MB^2\tag3$$

From $$(1)(2)(3)$$, we get $$BC'^2=C'M^2+MB^2\tag4$$ since \begin{align}&BC'^2-C'M^2-MB^2 \\\\&=(B'C'^2+BB'^2)-(B'C'^2+BB'^2-MB^2)-MB^2 \\\\&=0\end{align}

It follows from $$(4)$$ that $$C'M\perp A'B$$.

Similarly, we have $$A'N\perp BC'$$ and $$BP\perp A'C'$$.

In part a, you can think also in terms of vectors. For example, vector $$\vec{B'M}$$ is a linear combination of $$\vec i$$ and $$\vec k$$, but $$\vec{A'D'}$$ is a multiple of $$\vec j$$, assuming $$A'D'$$ lies on $$y$$-axis. So, $$A'D'\perp B'M.$$ Similarly for others.

In part b, you could also use Euclid's theorem for right triangles. For example for the right triangle $$\triangle BB'A'$$ with altitude $$B'M$$ to the hypotenuse $$BA'$$: $$BA'\,A'M=B'A'^2$$ $$BA'\,MB=B'B^2$$ Hence, $$\frac{A'M}{MB}=\frac{B'A'^2}{B'B^2}.\tag1$$ Similarly, $$\frac{BN}{NC'}=\frac{B'B^2}{B'C'^2}\tag2$$ and $$\frac{C'P}{PA'}=\frac{B'C'^2}{B'A'^2}\tag3$$ If we multiply the above equations and use reverse Ceva's theorem, we conclude that $$C'M, A'N, BP$$ are conccurent.