Let $ABCDA'B'C'D'$ be a right parallelepiped, $E$ and $F$ the projections of $A$ on the lines $A'D$, $A'C$, respectively, and $P, Q$ the projections of $B'$ on the lines $A'C'$ and $A'C$ Prove that a) the planes $(AEF)$ and $(B'PQ)$ are parallel b) the triangles $AEF$ and $B'PQ$ are similar.
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Tags: geometry, similar triangles, 3D geometry, parallelepiped
vanstraelen
19.08.2024 10:06
Is $ABCD$ a rectangle ?
rms
19.08.2024 10:15
vanstraelen wrote: Is $ABCD$ a rectangle ? Yes
vanstraelen
19.08.2024 19:41
Given the right parallelepiped $ABCDA'B'C'D'\ :\ A(0,0,0),B(b,0,0),C(b,d,0),D(0,d,0),A'(0,0,a),B'(b,0,a),C'(b,d,a),D'(0,d,a)$, $ABCD$ is a rectangle. Points $E(0,\frac{a^{2}d}{a^{2}+d^{2}},\frac{ad^{2}}{a^{2}+d^{2}})$ and $F(\frac{a^{2}b}{a^{2}+b^{2}+d^{2}},\frac{a^{2}d}{a^{2}+b^{2}+d^{2}},\frac{a(b^{2}+d^{2})}{a^{2}+b^{2}+d^{2}})$. Equation of the plane $(AEF)\ :\ bx+dy-az=0$. Points $P(\frac{b^{3}}{b^{2}+d^{2}},\frac{b^{2}d}{b^{2}+d^{2}},a)$ and $Q(\frac{b^{3}}{a^{2}+b^{2}+d^{2}},\frac{b^{2}d}{a^{2}+b^{2}+d^{2}},\frac{a(a^{2}+d^{2})}{a^{2}+b^{2}+d^{2}})$. Equation of the plane $(B'PQ)\ :\ bx+dy-az=b^{2}-a^{2}$. $\triangle AEF \sim \triangle B'PQ\ :\ \frac{AE}{B'P}=\frac{AF}{B'Q}=\frac{EF}{PQ}=\frac{a\sqrt{b^{2}+d^{2}}}{b\sqrt{a^{2}+d^{2}}}$.