Geometry Paper Folding Problem

GeometryTrianglesHardYKS

Published:

A piece of paper in the shape of triangle ABC is folded along segment $[AD]$, and after folding $[AC] // [DB']$.

$|AE| = 2 · |EB'|$ and $|BD| = 3 \text{ cm}$,

what is $|EC|$, in cm?

A) 5

B) 4

C) 6

D) $\frac{9}{2}$

E) $\frac{11}{2}$

This question includes visual content: A triangle ABC is illustrated. Part of it is folded along segment AD, resulting in a new vertex B'. The triangle is divided into a teal triangle (ADE) and a tan triangle (ADC). Segment BD is horizontal, and segment DB' is at an angle. The text specifies |AE| = 2|EB'| and |BD| = 3 cm, and mentions that after folding, [AC] is parallel to [DB'].

Animated Video Solution

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Step by Step Written Solution

1
Step 1

Hi Bengisu, let's solve this geometry folding problem together.

Geometry: Folding Triangle ABC

2
Step 2

When we fold triangle A B D along the segment A D to get triangle A B prime D, several properties stay the same.


Key Property of Folding:

Folding creates congruent triangles and angle bisectors.

3
Step 3

Because the folding line is A D, the angle B A D is equal to the angle B prime A D. This means A D is the angle bisector of angle B A B prime.

$$m(\angle BAD) = m(\angle B'AD) = \alpha$$
4
Step 4

The problem states that after folding, the segment A C is parallel to D B prime. Let's mark this key information.

$$AC \parallel DB'$$
5
Step 5

Since A C and D B prime are parallel, we can identify alternate interior angles. Specifically, angle C A D is equal to angle A D B prime. Let's call this measure beta.

$$m(\angle CAD) = m(\angle ADB') = \beta$$
6
Step 6

Let's look at the triangle properties we have gathered so far.

Analyzing the Triangle A E C

$$AE = 2 \cdot EB'$$
$$BD = 3 \implies DB' = 3$$
7
Step 7

Since B D is 3 and it was folded onto D B prime, the length of D B prime is also 3 centimeters.

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Step 8

Now consider the similarity between triangle D B prime E and triangle C A E because A C is parallel to D B prime.

By similarity (AA Criterion):

$$ \triangle DB'E \sim \triangle CAE$$
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Step 9

Because these triangles are similar, the ratio of their corresponding sides must be equal. This includes the ratio of D B prime to A C as well as the segments on the transversal lines.

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Step 10

The problem gives us the ratio for the segments on the transversal: A E equals two times E B prime.

$$\frac{EB'}{AE} = \frac{1}{2} $$
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Step 11

Substituting this back into our similarity ratio, we find that D B prime over A C must also be one over two.

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About This Question

Subject
Geometry
Topic
Triangles
Difficulty
Hard
Exam
YKS
Question Type
Multiple Choice

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