13.2 Parallelism and Perpendicularity Proofs in Solid Geometry

13.2 Parallelism and Perpendicularity Proofs in Solid Geometry

In the Gaokao Mathematics examination, Section I of solid geometry comprehensive questions almost universally demands a rigorous synthetic proof of parallelism or perpendicularity. Mastery of these formal proof theorems and their conversion logic is essential for securing full marks.

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1. Parallelism Proof Theorems

Line-to-Plane Parallelism (线面平行)

• Determination Theorem (判定定理): If a straight line $a$ outside a plane $\alpha$ is parallel to a straight line $b$ lying inside plane $\alpha$, then line $a$ is parallel to plane $\alpha$. $\begin{cases} a \not\subset \alpha \\ b \subset \alpha \\ a \parallel b \end{cases} \implies a \parallel \alpha$ Key Requirement: Must explicitly state $a \not\subset \alpha$ and $b \subset \alpha$.

• Property Theorem (性质定理): If a straight line $a$ is parallel to a plane $\alpha$, and a plane $\beta$ passing through $a$ intersects $\alpha$ along line $b$, then line $a$ is parallel to line $b$. $\begin{cases} a \parallel \alpha \\ a \subset \beta \\ \alpha \cap \beta = b \end{cases} \implies a \parallel b$

Plane-to-Plane Parallelism (面面平行)

• Determination Theorem (判定定理): If two intersecting straight lines $a$ and $b$ in plane $\alpha$ are both parallel to plane $\beta$, then plane $\alpha$ is parallel to plane $\beta$. $\begin{cases} a \subset \alpha, b \subset \alpha \\ a \cap b = P \\ a \parallel \beta, b \parallel \beta \end{cases} \implies \alpha \parallel \beta$ Key Requirement: The lines $a$ and $b$ MUST intersect ($a \cap b = P$).

• Property Theorem (性质定理): If two parallel planes $\alpha$ and $\beta$ are intersected by a third plane $\gamma$ along lines $a$ and $b$, respectively, then $a \parallel b$. $\begin{cases} \alpha \parallel \beta \\ \alpha \cap \gamma = a \\ \beta \cap \gamma = b \end{cases} \implies a \parallel b$

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2. Perpendicularity Proof Theorems

Line-to-Plane Perpendicularity (线面垂直)

• Determination Theorem (判定定理): If a straight line $l$ is perpendicular to two intersecting straight lines $a$ and $b$ lying in plane $\alpha$, then line $l$ is perpendicular to plane $\alpha$. $\begin{cases} a \subset \alpha, b \subset \alpha \\ a \cap b = O \\ l \perp a \\ l \perp b \end{cases} \implies l \perp \alpha$ Key Requirement: $a$ and $b$ must be inside $\alpha$ and must intersect.

• Property Theorem (性质定理): If two straight lines $a$ and $b$ are both perpendicular to the same plane $\alpha$, then $a$ is parallel to $b$. $\begin{cases} a \perp \alpha \\ b \perp \alpha \end{cases} \implies a \parallel b$ Definition Result: If $l \perp \alpha$, then $l \perp g$ for every line $g \subset \alpha$.

Plane-to-Plane Perpendicularity (面面垂直)

• Determination Theorem (判定定理): If a plane $\alpha$ contains a straight line $l$ that is perpendicular to a plane $\beta$, then plane $\alpha$ is perpendicular to plane $\beta$. $\begin{cases} l \subset \alpha \\ l \perp \beta \end{cases} \implies \alpha \perp \beta$

• Property Theorem (性质定理): If two planes $\alpha$ and $\beta$ are perpendicular to each other, and a straight line $a$ lying in plane $\alpha$ is perpendicular to their intersection line $l$, then line $a$ is perpendicular to plane $\beta$. $\begin{cases} \alpha \perp \beta \\ \alpha \cap \beta = l \\ a \subset \alpha \\ a \perp l \end{cases} \implies a \perp \beta$

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3. Logical Deductive Chain in Synthetic Proofs

Gaokao solid geometry proofs rely on translating geometric relationships back and forth across dimensions: $\text{Line } \parallel / \perp \text{ Line} \iff \text{Line } \parallel / \perp \text{ Plane} \iff \text{Plane } \parallel / \perp \text{ Plane}$

To prove plane $\alpha \perp \beta$:

1. Find an intersection line $l = \alpha \cap \beta$.
2. Locate or construct a line $a \subset \alpha$ such that $a \perp l$.
3. Prove $a \perp \text{another line } b \subset \beta$ intersecting $l$, yielding $a \perp \beta$.
4. Conclude $\alpha \perp \beta$ via the plane-to-plane perpendicularity determination theorem.

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4. Classic Gaokao Worked Example

Problem

As shown in the figure, in pyramid $P-ABCD$, the base $ABCD$ is a right isosceles trapezoid with $AB \parallel CD$, $\angle ABC = 90^\circ$, $AB = 2$, $CD = 1$, and $BC = \sqrt{2}$. Point $E$ is the midpoint of edge $PB$. Side face $PAD \perp$ base $ABCD$, and $PA = PD = \sqrt{3}$.

1. Prove line $EO \parallel$ plane $PDC$ (where $O$ is the midpoint of $BD$).
2. Prove plane $PAC \perp$ plane $PBD$.

Solution Part 1: Parallelism Proof

• Goal: Prove $EO \parallel$ plane $PDC$.
• Step 1: In $\triangle PBD$, $E$ is the midpoint of $PB$ and $O$ is the midpoint of $BD$.
• Step 2: By the triangle mid-line theorem, $EO \parallel PD$.
• Step 3: Check inclusion conditions: $PD \subset$ plane $PDC$, while $EO \not\subset$ plane $PDC$.
• Step 4: Applying the line-to-plane parallelism determination theorem: $\begin{cases} EO \not\subset \text{plane } PDC \\ PD \subset \text{plane } PDC \\ EO \parallel PD \end{cases} \implies EO \parallel \text{plane } PDC$

Solution Part 2: Perpendicularity Proof

• Goal: Prove plane $PAC \perp$ plane $PBD$.
• Step 1 (Base Geometry Analysis): In base trapezoid $ABCD$, set $B$ as the origin $(0,0)$. $BA$ lies along x-axis: $A=(2,0)$. $\angle ABC=90^\circ \implies C=(0, \sqrt{2})$. Since $CD \parallel AB$ and $CD=1$, $D=(1, \sqrt{2})$.
  • Vector $\vec{AC} = C - A = (-2, \sqrt{2})$. Vector $\vec{BD} = D - B = (1, \sqrt{2})$.
  • Dot product $\vec{AC} \cdot \vec{BD} = (-2)(1) + (\sqrt{2})(\sqrt{2}) = -2 + 2 = 0$.
  • Thus $\vec{AC} \perp \vec{BD} \implies AC \perp BD$.
• Step 2 (Pyramid Face Perpendicularity): Since $PA = PD = \sqrt{3}$, $\triangle PAD$ is an isosceles triangle. Let $M$ be the midpoint of $AD$. Then $PM \perp AD$.
• Step 3: Since face $PAD \perp$ base $ABCD$ and $PAD \cap ABCD = AD$, with $PM \subset PAD$ and $PM \perp AD$, by plane-to-plane perpendicularity property theorem, $PM \perp$ base $ABCD$.
• Step 4: Since $PM \perp$ base $ABCD$ and $BD \subset$ base $ABCD$, we have $PM \perp BD$.
• Step 5 (Line-Plane Perpendicularity): In plane $PAC$, line $BD$ is perpendicular to two intersecting lines $AC$ and $PM$ in plane $PAC$. $\begin{cases} AC \subset \text{plane } PAC, PM \subset \text{plane } PAC \\ AC \cap PM = H \\ BD \perp AC \\ BD \perp PM \end{cases} \implies BD \perp \text{plane } PAC$
• Step 6 (Plane-Plane Perpendicularity): Since $BD \subset$ plane $PBD$ and $BD \perp$ plane $PAC$, by the plane-to-plane perpendicularity determination theorem: $\text{plane } PAC \perp \text{plane } PBD$ (Q.E.D.)