From a sketch to the compass ; Geometric construction

[mathlearn]

Spoiler

In the traingle ABC, AB = 4.5cm , Ac = 9cm & angle ABC=90; The point D is located on BC such that DA = DC ;

Using only a straight edge with a cm/mm scale and a compass, construct an accurate diagram

Obtain the location of point E on side AB such that ACDE is a trapezium .

This is a rough sketch, (Happy)

View attachment 5924

Now apart from constructing the triangle can you help me to located the point D & Obtain the location of point E on side AB such that ACDE is a trapezium ,only using an straight edge and a compass. (Crying)

 

[I like Serena]

Spoiler

In the traingle ABC, AB = 4.5cm , Ac = 9cm & angle ABC=90; The point D is located on BC such that DA = DC ;

Using only a straight edge with a cm/mm scale and a compass, construct an accurate diagram

Obtain the location of point E on side AB such that ACDE is a trapezium .

This is a rough sketch, (Happy)
Now apart from constructing the triangle can you help me to located the point D & Obtain the location of point E on side AB such that ACDE is a trapezium ,only using an straight edge and a compass. (Crying)

Hey mathlearn! ;)

What can we say about triangle ACD?
What about its angles and sides?

 

[mathlearn]

Hey mathlearn! ;)

What can we say about triangle ACD?
What about its angles and sides?

(Nod)DA = DC \(\displaystyle \therefore\) It is an isosceles triangle.

(Happy) \(\displaystyle \angle DAC\) = \(\displaystyle \angle DCA\)

 

[I like Serena]

(Nod)DA = DC \(\displaystyle \therefore\) It is an isosceles triangle.

(Happy) \(\displaystyle \angle DAC\) = \(\displaystyle \angle DCA\)

Good! (Nod)

How about triangle ACM?

 

[mathlearn]

Good! (Nod)

How about triangle ACM?

(Nod)(Thinking) It's a 60 degree I guess.

 

[I like Serena]

(Nod)(Thinking) It's a 60 degree I guess.

Well, we'll have to get rid of the guessing, and also consider how to move forward. (Sweating)

Have you heard of congruence of triangles?
It means that they're the same, just positioned differently.

In particular triangles $\Delta ABC$ and $\Delta MBC$ are congruent.
Can you tell why?

Moreover, since that is the case, the angle at $C$ is divided into 2 equal angles.
So we can say something about the angles in triangle $\Delta ACD$! (Happy)

 

[mathlearn]

Well, we'll have to get rid of the guessing, and also consider how to move forward. (Sweating)

Have you heard of congruence of triangles?
It means that they're the same, just positioned differently.

In particular triangles $\Delta ABC$ and $\Delta MBC$ are congruent.
Can you tell why?

(Happy)

(Nod) $\Delta ABC$ \(\displaystyle \cong\) $\Delta MBC$ (SAS)

\(\displaystyle CB=CB\) (Common side)
\(\displaystyle \angle CBA= \angle CBM\) (CB perpendicular side AM)
\(\displaystyle AB=BM\) (data)

Moreover, since that is the case, the angle at $C$ is divided into 2 equal angles.
So we can say something about the angles in triangle $\Delta ACD$! (Happy)

\(\displaystyle \triangle ABC\) is half an equilateral triangle

(Nod) As one angle of an equilateral triangle is $60^\circ$; \(\displaystyle \therefore\) the angle at C is $30^\circ$

$\displaystyle \angle DAC = \displaystyle \angle DCA $So now as the problem states only to construct $\triangle BAC$ , From there I guess what you have to do is copy the angle $\angle DCA$ which is a $30^\circ$ at $\angle BAC$ and to draw a line through point A or the bisecting line of the angle. The point It meets the side BC is named D. Correct? (Thinking)

If so now to move on to constructing point E such that ACED is a trapezium (Sweating)(Crying)

 

[I like Serena]

(Nod) $\Delta ABC$ \(\displaystyle \cong\) $\Delta MBC$ (SAS)

\(\displaystyle CB=CB\) (Common side)
\(\displaystyle \angle CBA= \angle CBM\) (CB perpendicular side AM)
\(\displaystyle AB=BM\) (data)
\(\displaystyle \triangle ABC\) is half an equilateral triangle

(Nod) As one angle of an equilateral triangle is $60^\circ$; \(\displaystyle \therefore\) the angle at C is $30^\circ$

$\displaystyle \angle DAC = \displaystyle \angle DCA $So now as the problem states only to construct $\triangle BAC$ , From there I guess what you have to do is copy the angle $\angle DCA$ which is a $30^\circ$ at $\angle BAC$ and to draw a line through point A or the bisecting line of the angle. The point It meets the side BC is named D. Correct? (Thinking)

Good! (Nod)

However, I'm afraid that with a straight edge with a scale, and a compass, we can't just copy an angle. (Worried)
Instead we can do this in 2 ways.

With the compass we can divide any line segment into 2 equal parts, and draw a line perpendicular through it.
We can do this with $CM$ to find the line $AD$.

Alternatively, we can use trigonometry (sines, cosines, and tangents) to calculate $BD$ and measure it.

If so now to move on to constructing point E such that ACED is a trapezium (Sweating)(Crying)

Simplest is to use trigonometry to calculate $BE$...

 

[mathlearn]

Good! (Nod)

However, I'm afraid that with a straight edge with a scale, and a compass, we can't just copy an angle. (Worried)
Instead we can do this in 2 ways.

With the compass we can divide any line segment into 2 equal parts, and draw a line perpendicular through it.
We can do this with $CM$ to find the line $AD$.

Alternatively, we can use trigonometry (sines, cosines, and tangents) to calculate $BD$ and measure it.

Simplest is to use trigonometry to calculate $BE$...

(Nod) Thank you .

Now to Find the location of point E Such that ACDE is a trapezium a line parallel to CA through the point D should be constructed. Correct ? (Thinking)