prove

**yehoram** · August 8th 2010, 11:54 AM

Hi ! I have some problem with this below !

Subjected to a square which has two isosceles triangles.
One of them is: a triangle whose base is one of the sides of the square root angles are 15.
The other is a triangle whose base is the side opposite the base of the triangle it first touches the apex of the triangle at the end of the first node.
Proved that the second triangle equilateral!

The prove must be geometric way , no trigonometric ! Please help me ! thanks all !

**Vlasev** · August 8th 2010, 12:12 PM

You could begin by either describing the problem better or making a drawing.

**yehoram** · August 8th 2010, 10:12 PM

Here is a drawing of the problem !

**Soroban** · August 9th 2010, 03:13 AM

Hello, yehoram!

Code:

    E o - - - - - - - - o B
      |*              * |
      | *           *   |
      |  *        *     |
      |   *     *       |
      |    *  *         |
      |     o A         |
      |    *  *         |
      |   *     *       |
      |15*        *     |
      | *           *   |
      |*              * |
    D o - - - - - - - - o C

$\text{Given: }\,\text{square }BCDE.$
. . . . . . $\Delta BAE\text{ and }\Delta CAD\text{ are isosceles and congruent.}$
. . . . . . $\angle ADE = 15^o$

$\text{Prove: }\,\Delta ABC\text{ is equilateral.}$

$\angle ADE = 15^o \quad\Rightarrow\quad \angle ADC = 75^o$

$\text{Since }\Delta CAD\text{ is isosceles: }\:\angle DAC = 75^o$

$\text{In }\Delta CAD,\;\angle ACD = 30^o\quad\Rightarrow\quad \angle ACB = 60^o$

$\text{Since }\angle AED = 15^o\text{, we can show that: }\:\angle ABC = 60^o$

. . $\text{Hence: }\:\angle BAC = 60^o$

$\text{Therefore, }\Delta ABC\text{ is equilateral.}$

**yehoram** · August 9th 2010, 03:22 AM

Originally Posted by Soroban

Hello, yehoram!

$\angle ADE = 15^o \quad\Rightarrow\quad \angle ADC = 75^o$

$\text{Since }\Delta CAD\text{ is isosceles: }\:\angle DAC = 75^o$

$\text{In }\Delta CAD,\;\angle ACD = 30^o\quad\Rightarrow\quad \angle ACB = 60^o$

$\text{Since }\angle AED = 15^o\text{, we can show that: }\:\angle ABC = 60^o$

. . $\text{Hence: }\:\angle BAC = 60^o$

$\text{Therefore, }\Delta ABC\text{ is equilateral.}$

Who says that CAD is isosceles ?????

Subjected to just AED and ABC are isosceles !!!!!!!

**Soroban** · August 9th 2010, 04:19 AM

Originally Posted by yehoram

Who says that CAD is isosceles ?

Arrgh! . . . I read the problem incorrectly . . . *blush*

**Goethe** · August 9th 2010, 05:42 AM

angle DAE is 150 degrees, since EDA and DEA are 15 degrees each

because the triangles are inscribed in a square, the corners are right angles, and thus angles ADB and AEC are 75 degrees...assuming that triangle ABC is equilateral, all angles within the triangle are 60 degrees, which means that:

1. angle DAE+angle BAC is 150+60, which is 210...taking 210 from 360 degrees at point A, we find that the remaining 150 degrees are shared equally among angles DAB and EAC, meaning that both angles are 75 degrees...

2. angle ADB+angle DAB=150 degrees, meaning that angle ABD is 30 degrees

3. reflexively, angle CAE and angle AEC are 75 degrees, and angle ACE is 30 degrees

since all triangles equal 180 degrees, triangle ABC is in fact equilateral

I just joined the forum, and I apologize if

1. my answer is incorrect or isn't a rigorous proof

2. my answer is difficult to read, as i have no knowledge of LaTeX

**Vlasev** · August 9th 2010, 04:47 PM

Yehoram, where did you get this problem from? I'm asking because I think the problem is false.

**yehoram** · August 9th 2010, 07:22 PM

Originally Posted by Vlasev

Yehoram, where did you get this problem from? I'm asking because I think the problem is false.

Hi Vlasev ! Why did you think it's false . With trigonometric we can solve it and the answer is correct !!!

One of the teachers of my neighbor son gave them .

I heard that it have 4 ways of solution !!!!

**rainer** · August 10th 2010, 01:10 AM

With a compass you can demonstrate that a circle drawn from vertex DBC to DEC intersects the apex of the triangle DAE, thus proving that CA=CB, which would make ABC an equilateral.

**Vlasev** · August 10th 2010, 01:22 AM

I'm sorry, it's my bad. I calculated the angles with some trigonometry, but at a certain spot I assumed that an angle was 90 degrees when it wasn't, so in the end i was off by a few degrees. Now that i've checked it, the trig works.

**rainer** · August 10th 2010, 03:48 AM

Ok here's my attempt...

From angle EDA = 15 degrees we know angle ADB = 75 degrees. That's easy.

Now, draw a line DC from upper left corner to the lower right corner of the square. Because it's a square we know the angle CDE is 45 degrees. Call the intersection of DC with AB "P".

Using a compass draw a circle with center at the upper left corner of the square and with radius DA. You will find the circle intersects with P.

From angles CDE (45) and EDA (15) we can derive that angle ADP is 30 degrees. It is also clear that the circle radius DA=DP, meaning that triangle ADP is an isoceles. This means that the other two angles in triangle ADP must be 75 degrees.

Therefore angle DAB is 75 degrees. Since both DAB and ADB are 75 degrees, line AB must = DB, which means that ABC is an equilateral triangle.

In other words, triangle ABC is equilateral only for that angle EDA (which turns out to be 15) for which the above constructed circle intersects with P.

Does that work? I've never really tried to do something like this before.

**Vlasev** · August 10th 2010, 03:58 AM

That looks right if you can prove that the circle intersects at point P.

**sa-ri-ga-ma** · August 12th 2010, 10:21 PM

Draw MAN parallel to DB and PAQ parallel to DE. Join QM, MP, PN and NQ.

From the picture it is clear that angle AEM = EMQ = 15 degrees.

With center N and radius NQ draw an arc QRP, where P is on MA. Join RQ. Draw SRT parallel to ED.

According the the properties of the circle, angle SRQ = 1/2*angle RNQ.

Chord PQ = NQ when angle RNQ = 30 degrees or angle SRQ = 15 degrees.

But in the problem angle EMQ = 15 degrees.

Hence PQ = NQ when R coincides M or NQ = NM.

**yehoram** · August 13th 2010, 06:34 AM

Originally Posted by sa-ri-ga-ma

Draw MAN parallel to DB and PAQ parallel to DE. Join QM, MP, PN and NQ.

From the picture it is clear that angle AEM = EMQ = 15 degrees.

With center N and radius NQ draw an arc QRP, where P is on MA. Join RQ. Draw SRT parallel to ED.

According the the properties of the circle, angle SRQ = 1/2*angle RNQ.

Chord PQ = NQ when angle RNQ = 30 degrees or angle SRQ = 15 degrees.

But in the problem angle EMQ = 15 degrees.

Hence PQ = NQ when R coincides M or NQ = NM.

Can you picture it?? I don't undeastand your idea! thanks a lot !

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