Proving the Chord One Rule in Circle Theorems | Isosceles Triangle Proof

[adjacent]

One rule of circle theorems is,a line drawn from center to the mid-point of a chord cuts the cord at 90°.
What's the proof?
It's true that two radius and a chord creates an isosceles triangle.
So,

How can I prove that in an isosceles triangle,a line drawn from the vertex angle to the mid-point of the base side cuts the side at 90°?

 

[Nugatory]

You don't (yet) know that those two angles are 90 degrees... But what do you know about them?

 

[adjacent]

You don't (yet) know that those two angles are 90 degrees... But what do you know about them?

I'm not talking about the base anlgles.I am talking about the 90° angle you see in the image.
let vertex angle be y
$180-(\frac{y}{2}+x)=90$
$2x+y=180$

I can't seem to solve y and x.I don't know whether this is a proof or not.

 

[Nugatory]

I'm not talking about the base anlgles.
I am talking about the 90° angle you see in the image.

I'm sorry, I wasn't clear. Those two 90-degree angles are the two angles that I mean, and my question still stands: What do you know about them?

 

[adjacent]

I'm sorry, I wasn't clear. Those two 90-degree angles are the two angles that I mean, and my question still stands: What do you know about them?

hahahah.

I don't know what do you mean.
In a right angle triangle, $c^2=a^2+b^2$

c and b of both triangles is same.
so if it is two right angle triangles,a should be same
As the line was drawn to the midpoint,a is same in both triangles.So it is a right angle triangle.
Is this enough for a proof?

 

[Nugatory]

hahahah.

I don't know what do you mean.
In a right angle triangle, $c^2=a^2+b^2$

c and b of both triangles is same.
so if it is two right angle triangles,a should be same
As the line was drawn to the midpoint,a is same in both triangles.So it is a right angle triangle.
Is this enough for a proof?

No, because you've worked in the assumption both are right triangles, which is what you're trying to prove. The $c^2=a^2+b^2$ relationship isn't helping any because you don't know the values for all three to prove that you do have a right triangle...

But there's a reason I keep asking what you know about the two angles, not just one of them... What is their sum?

 

[adjacent]

No, because you've worked in the assumption both are right triangles, which is what you're trying to prove. The $c^2=a^2+b^2$ relationship isn't helping any because you don't know the values for all three to prove that you do have a right triangle...

But there's a reason I keep asking what you know about the two angles, not just one of them... What is their sum?

180°.Where any straight two lines intercept,any one side of the line has 180°.
but why should this help?If we have a 40° and a 50° angle,we would still get 180 as a sum.

 

[HallsofIvy]

Your midpoint line divides the large triangle into two smaller congruent triangles. They are congruent by "SSS": the two radii of the circle, the two bases, and the single ray in both triangles. From that follows that other "corresponding parts" are congruent.

 

[adjacent]

Your midpoint line divides the large triangle into two smaller congruent triangles. They are congruent by "SSS": the two radii of the circle, the two bases, and the single ray in both triangles. From that follows that other "corresponding parts" are congruent.

I know it's congruent.but how do I know that the angle is 90°?
I know that if vertex angle is y,the upper angle of both triangles have to be y/2

 

[Tanya Sharma]

Look at the attached figure.

Since ΔABD is congruent to ΔACD ,

∠ADB = ∠ADC

Now ∠ADB + ∠ADC =180° (Linear pair )

So, 2∠ADB = 180° or ∠ADB = 90°

 

[adjacent]

Look at the attached figure.

Since ΔABD is congruent to ΔACD ,

∠ADB = ∠ADC

Now ∠ADB + ∠ADC =180° (Linear pair )

So, 2∠ADB = 180° or ∠ADB = 90°

Thank you.Now that makes sense.
You all must be very skilled in proofs.I haven't even started.