# Triangle Problem: Proving DC=AC

• raul_l
In summary: Sorry.In summary, the conversation is discussing an attempt to prove a "theorem" that the length of a line segment is equal to the length of its subsegment. The individual asking for help has encountered a mistake in their calculations and is seeking assistance in identifying where the mistake has been made. After reviewing the equations, it is determined that the mistake is in the last equation, where it is assumed that DC and AC are equal, when in fact they are not.
raul_l
Hi.
What's wrong here?
I'm pretty sure I did all the operations correctly so I must be making some kind of conceptual flaw.

(first see the attachment)

$$\triangle ABC \sim \triangle BDC$$
$$\frac{S_{ABC}}{S_{BDC}}=\frac{AB^2}{BD^2}$$
$$\frac{S_{ABC}}{S_{BDC}}=\frac{\frac{AC \times BE}{2}}{\frac{DC \times BE}{2}}=\frac{AC}{DC} \Rightarrow \frac{AC}{DC}=\frac{AB^2}{BD^2}$$
$$AB^2=BE^2+AE^2=BC^2-EC^2+(AC-EC)^2=BC^2-EC^2+AC^2-2AC \times EC+EC^2=$$
$$=BC^2+AC^2-2AC \times EC$$
$$BD^2=BE^2+ED^2=BC^2-EC^2+(EC-DC)^2=BC^2-EC^2+EC^2-2EC \times DC+CD^2=$$
$$=BC^2-2EC \times DC+DC^2$$
$$\frac{AC}{DC}=\frac{BC^2+AC^2-2AC \times EC}{BC^2-2EC \times DC+DC^2}$$
$$\frac{BC^2-2EC \times DC+DC^2}{DC}=\frac{BC^2+AC^2-2AC \times EC}{AC}$$
$$\frac{BC^2}{DC}-2EC+DC=\frac{BC^2}{AC}+AC-2EC$$
$$\frac{BC^2}{DC}-AC=\frac{BC^2}{AC}-DC$$
$$\frac{BC^2-AC \times DC}{DC}=\frac{BC^2-AC \times DC}{AC} \Rightarrow DC=AC$$

#### Attachments

• att.JPG
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The first thing you are doing wrong is not defining your terms! What is SABC? The area of triangle ABC? I am willing to assume that A, B, C are vertices of triangle ABC but what is E? The foot of the altitude on AC? If BE is the length of the altitude of triangle ABC on AC, what right do you have to assert that is also the length of the altitude of triangle BDC on BD?

Erm.. I figured all those things were obvious enough.
So yes, S[abc] is the area of ABC just as S[bdc] is the area of BDC.
E is the foot of the altitude on AC and therefore BE is its length. However, it doesn't say that BE is the length of the altitude on BD. BE is the length of the altitude on DC, because it forms a right angle with the extension of DC (right?). Therefore 1/2xDCxBE is the area of triangle BDC (which can be seen in the 3rd line).
I now it's not much fun to look through these equations but I would really appreciate it if someone could tell me where I have made a mistake. Because otherwise I would be proving that the length of a line segment is equal to the length of it subsegment (not sure whether that's the right word), in this case that the length of AC equals the length of DC.

What are you actually trying to do?

0rthodontist said:
What are you actually trying to do?

Good question :)
Sorry I didn't make that clear.
Actually, finding the mistake is the only objective here. A friend of mine showed this to me. He said that he had proven the "theorem" that the length of a line segment is equal to the length of its subsegment (by using a triangle), which was obviously a joke. However, he showed me how he had done that and asked me to find a mistake there. And so far I've got nothing.
If you look at the equations then they're all very simple operations, basically I've only used the Pythagorean theorem and a property of similar triangles. So to answer you question, the objective is to prove this "theorem" wrong.

I just found the answer to this problem.
In the last equation $$BC^2-AC \times DC=0$$ and therefore it would be incorrect to assume that DC=AC.

raul_l said:
Erm.. I figured all those things were obvious enough.
So yes, S[abc] is the area of ABC just as S[bdc] is the area of BDC.
E is the foot of the altitude on AC and therefore BE is its length. However, it doesn't say that BE is the length of the altitude on BD. BE is the length of the altitude on DC, because it forms a right angle with the extension of DC (right?). Therefore 1/2xDCxBE is the area of triangle BDC (which can be seen in the 3rd line).
I now it's not much fun to look through these equations but I would really appreciate it if someone could tell me where I have made a mistake. Because otherwise I would be proving that the length of a line segment is equal to the length of it subsegment (not sure whether that's the right word), in this case that the length of AC equals the length of DC.

No, (1/2)(DC x BE) does not equal the area of triangle BDC. The height must be a line drawn to the vertex of a triangle.

EDIT: Nevermind. I just proved that I was wrong.

Last edited:

## 1. What is the Triangle Problem: Proving DC=AC?

The Triangle Problem: Proving DC=AC is a mathematical problem that involves proving that the length of side DC of a triangle is equal to the length of side AC.

## 2. Why is proving DC=AC important?

Proving DC=AC is important because it is a fundamental concept in geometry and it allows us to understand the relationships between the sides and angles of a triangle. It also serves as a building block for more complex geometric proofs.

## 3. What is required to prove DC=AC?

To prove DC=AC, we need to use a combination of known geometric theorems and properties, such as the Pythagorean theorem, congruent triangles, and angle relationships. We also need to use logical reasoning and deductive arguments to prove the statement.

## 4. How can we prove DC=AC?

One way to prove DC=AC is by using the Side-Side-Side (SSS) congruence theorem, which states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. By showing that the two triangles formed by DC and AC are congruent, we can prove that their corresponding sides, DC and AC, are equal in length.

## 5. Are there any real-life applications of proving DC=AC?

Yes, proving DC=AC has real-life applications in fields such as architecture, engineering, and construction. In these fields, understanding the relationships between the sides and angles of a triangle is crucial for designing and building structures that are stable and structurally sound.

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