# Show that angle AXC=angle ACB; geometric construction

• MHB
• mathlearn
In summary,Looks like this question is going to make a long thread. :)This is what the problem states:Using the pair of compass and a straight edge,(i) Construct the triangle ABC according to the measurements shown in the given sketch.Then,(ii) Construct a perpendicular from A to BC and name the point it meets BC as D. After That,(iii) Construct the circle that passes through the points A, C and D. Thereafter,(iv) Construct the tangent to this circle at the point C, and name the point at which it meets AD produced as X.
mathlearn
Looks like this question is going to make a long thread. :) This is what the problem states

"Use only a straight edge with a cm/mm scale and a pair of compasses to do the following constructions. Draw your construction lines clearly.

View attachment 5914

(i) Construct the triangle ABC according to the measurements shown in the given sketch.

(ii) Construct a perpendicular from A to BC and name the point it meets BC as D.

(iii) Construct the circle that passes through the points A, C and D.

(iv) Construct the tangent to this circle at the point C, and name the point at which it meets AD produced as X.

(v) Show that $\angle$AXC = $\angle$ACB."

Using the pair of compass and a straight edge,
(i) Construct the triangle ABC according to the measurements shown in the given sketch.

Then,
(ii) Construct a perpendicular from A to BC and name the point it meets BC as D.

After That,
(iii) Construct the circle that passes through the points A, C and D.

Thereafter,
(iv) Construct the tangent to this circle at the point C, and name the point at which it meets AD produced as X.

Thereafter, The question states to mark the point where the tangent and AD produced as 'X'.Now can you help me to state, $\angle$AXC = $\angle$ACB. and I want to know verify the rest is done correctly. :) The updated diagram,

(v) Show that $\angle$AXC = $\angle$ACB.

Many thanks :)

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mathlearn said:
Now can you help me to state, $\angle AXC = \angle ACB$.
You have just stated it. Do you need to prove it? Note that a circumcircle around a right triangle has the hypotenuse as a diameter; therefore, $XC\perp AC$.

Yes the reasons must be given for the angles to be equal .Not getting It so far , An explanation with little detail would help :)

$$\angle{ACD}=90^\circ-\angle{CAD}$$

$$\angle{AXC}=\,?$$

greg1313 said:
$$\angle{ACD}=90^\circ-\angle{CAD}$$

$$\angle{AXC}=\,?$$

Hmm. True but the problem states To give reasons why $\angle AXC = \angle ACB$, not $\angle{ACD}$

As the given sketch looks a little bit difficult to refer, This sketch would help

View attachment 5917

This sketch looks easy to refer , and do you notice how $\angle AXC = \angle ACB$, in the sketch ; Now help me to state the reasons , please do your best to give a detailed reply really quick :)

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mathlearn said:
True but the problem states To give reasons why $\angle AXC = \angle ACB$, not $\angle{ACD}$
What's the difference?

You've been told that $\angle{ACD}=90^\circ-\angle{CAD}$. It is left to note that $\angle{AXC}=90^\circ-\angle{CAD}$ because $AC\perp CX$.

mathlearn said:
Hmm. True but the problem states To give reasons why $\angle AXC = \angle ACB$, not $\angle{ACD}$

My apologies, (Smile)

greg1313 said:
$$\angle{ACD}=90^\circ-\angle{CAD}$$

$$\angle{AXC}=\,?$$

Evgeny.Makarov said:
You've been told that $\angle{ACD}=90^\circ-\angle{CAD}$. It is left to note that $\angle{AXC}=90^\circ-\angle{CAD}$ because $AC\perp CX$.

Oh! consider the two $$\displaystyle \triangle$$ ADC & $$\displaystyle \triangle$$ AXC

So that,
Evgeny.Makarov said:
You've been told that $\angle{ACD}=90^\circ-\angle{CAD}$. It is left to note that $\angle{AXC}=90^\circ-\angle{CAD}$ because $AC\perp CX$.

View attachment 5918

Am I correct ? :) What theorem was used to gain this approach?

Many Thanks :)

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You have not written any reasoning except "Consider two triangles", so it's difficult to say whether you are correct.

Also please note that the forum rules prohibit asking for help in private messages.

Oh! I'm sorry (Doh)

Evgeny.Makarov said:
You have not written any reasoning except "Consider two triangles", so it's difficult to say whether you are correct.

So considering the two triangles ADC & ACX,

$$\displaystyle \angle{ACD}=90^\circ-\angle{CAD}$$ & $$\displaystyle \angle{AXC}=90^\circ-\angle{CAD}$$

(v) Show that ∠AXC = ∠ACB.

$$\displaystyle \therefore$$ $$\displaystyle \angle{ACD}=\angle{AXC}$$

And Is there any theorem being used here?

Many Thanks (Smile)

mathlearn said:
$$\displaystyle \angle{ACD}=90^\circ-\angle{CAD}$$ & $$\displaystyle \angle{AXC}=90^\circ-\angle{CAD}$$

$$\displaystyle \therefore$$ $$\displaystyle \angle{ACD}=\angle{AXC}$$

And Is there any theorem being used here?
Yes. The theorem that is used here says: For all numbers $x$, $y$ and $z$, if $x=z$ and $y=z$, then $x=y$. It's a basic property of equality.

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## 1. How can I prove that angle AXC is equal to angle ACB using geometric construction?

To prove that angle AXC is equal to angle ACB, you can use the following steps in geometric construction:

• Draw line segment AB and point C on the line.
• Construct a circle centered at point C with radius CB.
• Draw a line from point A that intersects the circle at points X and Y.
• Construct a line perpendicular to line segment AB passing through point C.
• Draw a line from point X that intersects the perpendicular line at point Z.
• Using the compass, construct a circle centered at point Z that passes through points X and Y.
• The intersection point of the two circles, point X, is the vertex of angle AXC. The two angles formed by this vertex, AXC and ACB, are equal.

## 2. Can I use a protractor to prove that angle AXC is equal to angle ACB?

No, you cannot use a protractor to prove that angle AXC is equal to angle ACB. Geometric construction involves using only a straightedge and a compass, not a protractor. The steps outlined above show how to prove equal angles using geometric construction.

## 3. How does geometric construction help in proving equal angles?

Geometric construction is a visual method for proving geometric theorems. By constructing lines, circles, and angles using only a straightedge and a compass, we can visually see the relationships between different geometric elements, such as angles. This makes it easier to prove that two angles are equal.

## 4. Are there any other methods to prove that angle AXC is equal to angle ACB?

Yes, there are other methods to prove that angle AXC is equal to angle ACB, such as using algebraic equations and trigonometric identities. However, geometric construction is a more visual and intuitive method for proving equal angles.

## 5. Can we use geometric construction to prove that two angles are not equal?

Yes, we can use geometric construction to prove that two angles are not equal. If the construction steps do not result in the two angles having the same vertex and size, then we can conclude that they are not equal. Additionally, we can also use geometric construction to construct a specific angle and compare it to the given angle to determine if they are equal or not.

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