Prove that triangles $VAC$ and $VBD$ have equal areas and equal perimeters....

Question

The question

Let $VABCD$ be a quadrilateral pyramid with a rectangular base. $\angle AVC =\angle BVD$ prove that triangles $VAC$ and $VBD$ have equal areas and equal perimeters.

The idea

Because the base ABCD is rectangular we get that $DB=AC$.

We also know the congruences of the angles, so I was thinking of showing that VDB is congruent with VDC this will make the perimeters and the areas equal.

I don't know how to show this. I hope one of you can help me! Thank you!

Answer 1

As commented by Blue, the claim is not true in general.

In the following, I'm going to prove the following claim :

Claim 1 : If $\angle AVC =\angle BVD\color{red}{\not=90^\circ}$, then triangles $VAC$ and $VBD$ have equal areas and equal perimeters.

Proof :

Applying the law of cosines to $\triangle{AVC}$ and $\triangle{BVD}$, we have $$\cos\angle{AVC}=\frac{VA^2+VC^2-AC^2}{2\times VA\times VC}\tag1$$ $$\cos\angle{BVD}=\frac{VB^2+VD^2-BD^2}{2\times VB\times VD}\tag2$$

Here, let $E$ be a point on the plane $ABCD$ such that $EV\perp ABCD$.

Let $F$ be a point on $AB$ such that $EF\perp AB$.

Let $G$ be a point on $CD$ such that $EG\perp CD$.

Here, we have $$\begin{align}&VA^2+VC^2-AC^2 \\\\&=VE^2+EA^2+VE^2+EC^2-AC^2 \\\\&=2VE^2+EF^2+FA^2+EG^2+GC^2-AC^2\end{align}$$ and $$\begin{align}&VB^2+VD^2-BD^2 \\\\&=VE^2+EB^2+VE^2+ED^2-BD^2 \\\\&=2VE^2+EF^2+FB^2+EG^2+GD^2-BD^2\end{align}$$

Since $FA=GD,GC=FB,AC=BD$, we have $$VA^2+VC^2-AC^2=VB^2+VD^2-BD^2\not=0$$

So, from $(1)(2)$, we get $$\frac{1}{2\times VA\times VC}=\frac{1}{2\times VB\times VD}$$ which implies $$\begin{align}&VA^2\times VC^2=VB^2\times VD^2 \\\\&\implies (VE^2+EF^2+FA^2)(VE^2+EG^2+GC^2)=(VE^2+EF^2+FB^2)(VE^2+EG^2+GD^2) \\\\&\implies (VE^2+EF^2+FA^2)(VE^2+EG^2+GC^2)=(VE^2+EF^2+GC^2)(VE^2+EG^2+FA^2) \\\\&\implies EF^2GC^2+FA^2EG^2=EF^2FA^2+GC^2EG^2 \\\\&\implies (GC^2-FA^2)(EF^2-EG^2)=0 \\\\&\implies (GC-FA)(EF-EG)=0\end{align}$$

• $GC=FA$ implies that $V$ is on the plane which passes through the midpoints of $AB,CD$, and is perpendicular to the plane $ABCD$. Then, since $VA=VB$ and $VC=VD$, we can say that $\triangle{VAC}\equiv\triangle{VBD}$.

• $EF=EG$ implies that $V$ is on the plane which passes through the midpoints of $AD,BC$, and is perpendicular to the plane $ABCD$. Then, since $VA=VD$ and $VC=VB$, we can say that $\triangle{VAC}\equiv\triangle{VDB}$.

Therefore, the claim 1 follows.$\ \blacksquare$

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Added :

May I ask why the case where the angles equal 90 can't conclude our desired demonstration?

The following claim is false :

Claim 2 : If $\angle AVC =\angle BVD=90^\circ$, then triangles $VAC$ and $VBD$ have equal areas and equal perimeters.

Blue has already given an explanation in a comment. I'll try to add some explanations.

Let $H$ be the intersection point of $AC$ with $BD$.

To find a counterexample of Claim 2, let us consider a sphere $S$ whose center is $H$ with radius $HA$.

Note that the four points $A,B,C,D$ are on $S$.

For any point $P$ (except $A,B,C,D$) on $S$, we have $\angle APC =\angle BPD=90^\circ$.

Now, as a counterexample of Claim 2, let us consider a pyramid $VABCD$ where $V$ is on $S$ and is very close to $A$.

Since $V$ is on $S$, we have $\angle AVC =\angle BVD=90^\circ$.

• The area of $\triangle{VAC}$ is approximately $0$ while the area of $\triangle{VBD}$ is approximately the area of $ABD$. So, triangles $VAC$ and $VBD$ do not have equal areas.

• The perimeter of $\triangle{VAC}$ is approximately $2AC$ while the perimeter of $\triangle{VBD}$ is approximately $AB+AD+BD$, so triangles $VAC$ and $VBD$ do not have equal perimeters.

Therefore, such a pyramid $VABCD$ is a counterexample of Claim 2.

Answer 2

Let $A(0,0,0),$ $B(0,b,0),$ $C(a,b,0),$ $D(a,0,0)$ and $V(x,y,z)$. Then, the perimeter condition $VA+VD=VB+VC$ and the area condition $\Vert \vec{VA}\times\vec{VD}\Vert=\Vert\vec{VB}\times\vec{VC}\Vert$ give $$\sqrt{x^2+y^2+z^2}+\sqrt{(x-a)^2+y^2+z^2}=\sqrt{x^2+(y-b)^2+z^2}+\sqrt{(x-a)^2+(y-b)^2+z^2}\tag1$$ and $$y=\frac b2.\tag2$$ If $y=\frac b2$ then $(1)$ is satisfied. So, for counter-example we must choose $y\neq\frac b2.$

Blue already gave counter-examples in comments. My counter-example: $a=2,b=1, x=0.5, y=0.4, z\approx 0.498422$ where $z$ is a solution of $$\frac{x^2+y^2+z^2-ax}{\sqrt{x^2+y^2+z^2}\sqrt{(x-a)^2+y^2+z^2}}=\frac{x^2+(y-b)^2+z^2-ax}{\sqrt{x^2+(y-b)^2+z^2}\sqrt{(x-a)^2+(y-b)^2+z^2}}$$ obtained by the law of cosines as Mathlove did.

Answer 3

Intuitively,

Consider (an easier 3d development) of tetrahedron VABCD. Figure at left:

Draw identical circles separately through $ (AD,BC)$ whose line of centers is parallel to AB or CD. Note the altitudes from $(V1,V2)$ to AD or BC are equal. Also note that the angles subtended at $(V1,V2)$ inside identical circle arcs are equa, by virtue of symmetry with respect to axis Ox.

Now fold the triangles $ (V1DA, V2CB) along fold/hinge lines (DA,CB) inward until V1, V2 come together at V, call it vertex V. The symmetry (coming from identical circles) forces Ox to bifurcate sides AB and CD by the plane OxV.

Also note angles contained in circular segments DVA,CBB in those inclined triangles must be equal.

For lines in plane of ABCD there is symmetry about a central horizontal mirror x- axis, but not about the y-axis. Altitude $h$ is perpendicular the plane of ABCD.

Triangles (VAC, VBD) are congruent and it proves as required..

By projection of sides in the unsymmetrical tetrahedron

Vertex V must be contained in a plane bisecting rectangle ABCD and is perpendicular to the plane of this rectangle. We choose V directly above O.

Let $\angle XOV$ be $90^{\circ}$although wlog it can be any angle either acute or obtuse.

Coordinates ABCD: $$ (c,b,0),(c,-b,0), (-a,-b,0),(-a,b,0)~ $$

Coordinates of V (directly above O): $$ (0,0,h)$$ Since sides $(a,b,h) $ are along $(x,y,z)$ directions $$ VC=VD=\sqrt{ a^2+b^2+h^2} $$ Similarly $$ VA=VB=\sqrt{ c^2+b^2+h^2} $$

Triangles $(VCA,VDB)$ are congruent $ ( S,S,S ) $ congruence.

Correspondingly the triangles $(AVC,BVD)$ are congruent and corresponding angles $(CVA,DVB)$ are also equal.

In the above we have Ox x-axis equidistant/parallel between AD and BC choosing O on Ox not same as intersection center point of diagonals AC,BD to show congruence.

Exactly in the same way we can have or draw Oy y-axis equidistant/parallel between AB and CD choosing O on Oy not same as intersection center point of diagonals AC,BD to show the same congruence.

The congruence shows equality of not just perimeters and areas, but all associated triangle properties like the In-Radius, Ex-Radii, Circum-Radius, Nine Point Circle-Radius etc.

Answer 4

Define $$a:=|AV| \quad b := |BV| \quad c := |CV| \quad d := |DV| \\[4pt] s := |AC|=|BD|$$ Note that the desired equal-perimeter property for $\triangle AVC$ and $\triangle BVD$ means $$a+c+s = b+d+s \qquad\to\qquad a+c = b+d \tag1$$ while the desired equal-area property, along with the given equal-angle property, means $$ \tfrac12ac\sin\angle AVC = \tfrac12 b d\sin\angle BVD \qquad\to\qquad ac = bd \tag2$$ Writing $p$ for the sum in $(1)$, and $q$ for the product in $(2)$, we see that $a$ and $c$ are the roots of $x^2-px+q=0$; but so are $b$ and $d$! We conclude, then, that we require either

• $a=b$ and $c=d$: $V$ lies on the perpendicular-bisector plane of $AB$ (and $CD$).
• $a=d$ and $b=c$. $V$ lies on the perpendicular-bisector plane of $AD$ (and $BC$).

Note that these cases make the triangles fully-congruent (by Side-Side-Side). Any point satisfying them provides the equal-angle property, as well; there are no "extraneous" solutions.

However, as I mentioned in a comment, the equal-angle property is satisfied by any point on the sphere of radius $s/2$ centered at the rectangle's center. Thus, the equal-angle property itself does not guarantee the equal-perimeter and equal-area properties. This makes the original problem statement incorrect.

(Given that the only way this all works is for the triangles to be fully-congruent, separating-out the equal-perimeter and equal-area properties seems rather odd. I'm curious what the author of the original question was thinking.)

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Addendum. To flesh-out what the equal-angle property alone does and doesn't provide, let's coordinatize, with $$A=(h,k) \quad B =(-h,k) \quad C=(-h,-k) \quad D=(h,-k) \quad V = (u,v,w)$$ and $a$, $b$, $c$, $d$ as above.

We can calculate $$\begin{align} \cos\angle AVC &= \frac{(A-V)\cdot(C-V)}{ac} = \frac{(u^2+v^2+w^2)-(h^2+k^2)}{ac}=\frac{|OV|^2-|OA|^2}{ac} \\[4pt] \cos\angle BVD &= \frac{(B-V)\cdot(C-V)}{bd} = \cdots = \frac{|OV|^2-|OA|^2}{bd} \end{align}$$ The equal-angle property implies that the cosines are equal, and we can write $$(ac-bd)\left(|OV|^2-|OA|^2\right) = 0 \tag3$$

• If $ac-bd=0$, then also $$0=a^2c^2-b^2d^2=\|A-V\|^2\|C-V\|^2-\|B-V\|^2\|D-V\|^2=-16hkuv \tag4$$ From here, $u=0$ or $v=0$ implies that $V$ lies on either the $yz$- or $xz$-plane (ie, the perpendicular-bisector plane of $AB$ and $CD$, or of $AD$ and $BC$). As before, this makes the triangles fully-congruent, with equal areas and perimeters.

• If $|OV|^2-|OA|^2=0$ (which corresponds to $\angle AVC=\angle BVD=90^\circ$), then $V$ is on the origin-centered sphere of radius $r:=s/2$. The many points of this sphere that are not also on the $yz$- or $xz$-planes cannot satisfy the equal-perimeter and equal-area properties together, so this case does not guarantee the desired conclusion.

So, the original problem statement could be fixed by specifying that $\angle AVC$ and $\angle BVD$ are non-right ... although that scenario would weirdly exclude the points where the sphere meets the $yz$- and $xz$-planes from set of allowable locations for $V$.