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The street lamps are $15$ and $10$ feet tall. How tall is the Grim Reaper?

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Move the point on the side of the triangle. The applet draws a line parallel to another side of the triangle and gives the length of four line segments.

- Move the vertices and investigate the relationship between these lengths. What do you notice?
- Explore different triangles in relation to the pyramids!

The converse of the **Side-Splitter Theorem** is also true.

If a segment is drawn between two sides of a triangle such that it divides the sides proportionally, the segment is parallel to the third side in the triangle.

Based on the diagram, the following relation holds true.

If $DCAD =ECBE ,$ then $DE∥AB.$

The given proportion can be rearranged to get the proportionality of two sides of $△ABC$ and $△DEC.$
This means that $△ABC$ is a dilation of $△DEC$ from point $C$ with scale factor $r=DCAC =ECBC .$
A dilation moves a segment to a parallel segment, so the proof is complete.

$DCAD =ECBE $

$DCAC =ECBC $

If $DCAD =ECBE ,$ then $DE∥AB.$

Show that $PQRS$ is a parallelogram.

Draw the diagonals of quadrilateral $ABCD.$

Draw diagonal $AC$ of quadrilateral $ABCD$ and focus on the two triangles $△ABC$ and $△ADC.$

The given measures of the segments make it possible to derive the ratios according to how the transversal $PQ $ divides sides $AB$ and $BC$ of $△ABC.$$PBPA QBQC =1.84.2 =37 =2.45.6 =37 $

It can be seen that the two ratios are equal. Therefore, according to the converse of the Triangle Proportionality Theorem, the transversal $PQ $ is parallel to the diagonal $AC.$ A similar calculation shows that $SR$ cuts the sides of $△ADC$ proportionally. Therefore, $SR$ is also parallel to the diagonal $AC.$
$SDSA RDRC =2.14.9 =37 =2.76.3 =37 $

Since $PQ $ and $SR$ are both parallel to $AC, $ they are also parallel to each other.
After drawing the diagonal $BD,$ the previously found proportions also show that $PS$ cuts the sides $AB$ and $AD$ of $△ABD$ proportionally. Additionally, $QR $ cuts the sides $CB$ and $CD$ of $△CBD$ proportionally.
$PBPA QBQC =SDSA =37 =RDRC =37 $

This implies that $PS$ and $QR $ are both parallel to $BD.$ Therefore, they are also parallel to each other.
This completes the proof that opposite sides of quadrilateral $PQRS$ are parallel. Hence, by definition, it is a parallelogram.

The following theorem is a corollary of the Side-Splitter Theorem.

Construct points that divide the given segment into five congruent pieces.

Draw a different segment and extend it with four congruent copies.

Draw a ray starting at $A$ and use a compass to copy any length five times on this ray. This gives five points, $P_{1},$ $P_{2},$ $P_{3},$ $P_{4},$ and $P_{5}.$

Connect $B$ with the last point, $P_{5},$ and construct parallel lines to this segment through the other points. Mark the intersection points of these lines with segment $AB.$

According to the Three Parallel Lines Theorem, these transversals divide segments $AB$ and $AP_{5} $ proportionally. Since, by construction, the segments on $AP_{5} $ have equal length, this means that points $Q_{1},$ $Q_{2},$ $Q_{3},$ and $Q_{4}$ divide $AB$ into congruent segments.

The previous exploration can lead to the following claim.

The following two claims are corollaries of the Right Triangle Similarity Theorem

To conclude this lesson, the opening challenge will be revisited. The challenge shows a diagram consisting of the Grim Reaper and two street lamps at $15$ and $10$ feet tall. How tall is the Grim Reaper?

The lamps, the head of the Grim Reaper, and the shadows of the Grim Reaper's head are on a straight line.

The lamps and the figure stand vertically. Hence, they can be represented by parallel segments $AB,$ $CD,$ and $EF$. Notice that the shadows just reach the lampposts. Taking a look at the shadow touching the taller post, it can be derived that the lamppost bottom $B,$ the head of the Grim Reaper $C,$ and the lamppost top $E$ are on a straight line. Applying this same logic to the other shadow implies that $A,$ $C,$ and $F$ are also on a straight line.

Segment $CD$ splits both $△ABF$ and $△BEF.$ It is also parallel to one side of both triangles. That means the combination of two dilations maps $AB$ to $EF.$

- A dilation with center $F$ and scale factor $DF:BF$ maps $AB$ to $CD.$
- A dilation with center $B$ and scale factor $BF:BD$ maps $CD$ to $EF.$

Continuing, notice that Point $D$ is between $B$ and $F.$ Therefore, the Segment Addition Postulate guarantees that $BD+DF=BF.$ Then, to divide this equality by $BF$ and to rearrange it gives a relationship between the scale factors of the two dilations.

$BD+DF=BF$

$BDBF =1−BFDF 1 $

$BFDF ⋅BDBF =1510 $

SubstituteExpressions

Substitute expressions

$BFDF ⋅1−BFDF 1 =1510 $

Solve for $BFDF $

Substitute

$BFDF =x$

$x⋅1−x1 =1510 $

MultEqn

$LHS⋅15(1−x)=RHS⋅15(1−x)$

$15x=10(1−x)$

Distr

Distribute $10$

$15x=10−10x$

AddEqn

$LHS+10x=RHS+10x$

$25x=10$

DivEqn

$LHS/25=RHS/25$

$x=2510 $

CalcQuot

Calculate quotient

$x=0.4$

Substitute

$x=BFDF $

$BFDF =0.4$

$CD=0.4(15)=6 $

Segment $CD$ represents the Grim Reaper, who is now known to stand at $6$ feet tall.