Help With Euclidean Geo. Question in Cambridge Yr.12 (1 Viewer)

414076 · Jan 28, 2016

Having issues with Q(7)(b) in Cambridge Yr. 12 3U 8F (Areas of Plane Figures). The answers at the back say that "triangles with the same base and area have the same altitude", which means that the triangles would have the same height. But I don't see how this would lead to AB being // with DC? Help would be appreciated.

InteGrand · Jan 28, 2016

414076 said:
Having issues with Q(7)(b) in Cambridge Yr. 12 3U 8F (Areas of Plane Figures). The answers at the back say that "triangles with the same base and area have the same altitude", which means that the triangles would have the same height. But I don't see how this would lead to AB being // with DC? Help would be appreciated.

View attachment 32835

$\noindent Well basically, since $\triangle ABD$ and $\triangle ABC$ have a common base $AB$ and have equal area, their height is the same, as you said the answers says. This means that $D$ has to be the same perpendicular height above $AB$ as $C$, and this forces $AB\parallel DC$. This can be proved as follows.$

$\noindent In our diagram, let $D^\prime$ and $C^\prime$ be the foot of the perpendicular respectively from $D$ and $C$ to $AB$. Then $CC^\prime = DD^\prime$, as we said the two triangles had equal height. Also, $\angle DD^\prime C^\prime = 90^\circ = \angle C C^\prime D^\prime$ (definition of perpendicular). Hence $DD^\prime \parallel CC^\prime$ (cointerior angles are supplementary). Hence quadrilateral $D^\prime C^\prime CD$ has a pair of sides that are both equal and parallel (namely $DD^\prime$ and $CC^\prime$), which means it is a parallelogram (this is based on the fact that a quadrilateral is a parallelogram if and only if it contains a pair of sides that are both equal and parallel). Hence $D^\prime C^\prime \parallel DC$. But $D^\prime C^\prime$ lies on $AB$. So $AB\parallel DC$. Q.E.D.$

leehuan · Jan 28, 2016

Keyword: Basically

InteGrand · Jan 28, 2016

leehuan said:
Keyword: Basically

What I meant by that was that you can "see it" (that the lines have to be parallel because the two points are equal height above a given line). The OP phrased it as though they couldn't see why the lines were parallel, so I provided the intuition for it before providing a proof.

Another easy way to show the result is to construct an x-y coordinate system with the x-axis on AB and the y-axis perpendicular to this. Then since the points C and D lie at equal heights above AB (the x-axis), they both lie on the same horizontal line y = h, say (where h is the height they are above AB). So CD is a horizontal line with respect to the x-axis, as is the x-axis of course, so AB is parallel to CD.

leehuan · Jan 28, 2016

InteGrand said:
What I meant by that was that you can "see it" (that the lines have to be parallel because the two points are equal height above a given line). The OP phrased it as though they couldn't see why the lines were parallel, so I provided the intuition for it before providing a proof.

Yeah. I know. But for some reason I felt like that called for a tease

Help With Euclidean Geo. Question in Cambridge Yr.12 (1 Viewer)

414076

New Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)