Solution-verification | Prove that the lines $BP,A’N,C’M$ are concurrent

geometrysolution-verification

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

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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!

Best Answer

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 $BP\perp A'C'$.

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