# Question: Circle with Parallel Lines

## Comment on Circle with Parallel Lines

### We also can find that AC=BD,

We also can find that AC=BD, since AB//CB, so <BAD=<ADC=10 degree, which is the half of <BOD. <ECD=90 degree, <ADC=10, so X=80. Perfect!!

### it will fit for me.

it will fit for me.

### Please let me know if this

Please let me know if this works even in the case of the figure not being drawn to scale:

As the line ED passes through parallel lines, we know that <EDC must = 20. If we can confirm that line AD does exist, then <ADC must be less than 20 since it splits somewhere between ED and CD. We also know that <ECD = 90 as shown in the video. With this, the only answer that will add to 180 when X + 90 + (some number less than 20) is combined is 80 (E) ### Very nice - I love it!!!!

Very nice - I love it!!!!

### Hi,

Hi,
In the question why can't we consider angle BOD and angle AOE as opposite as they clearly appear so. If i take it as opposite angle E = 180-90+20 = 70. Angle E and angle x are equal as they both have common chord CD. Hence, x=70. Please explain me how i am wrong (apparently in consideration of opposite angle only). ### Be careful.

Be careful.

Yes, angle BOD and angle AOE are opposite angles, which means they are equal.

However, we cannot use the rule that says inscribed angles holding the same chord are equal, because angle x is not an inscribed angle. It is not an inscribed angle, because it is not ON the circle.

### In this Question, O has 20*

In this Question, O has 20* why not opposite angles are equaland then two opposite angle equals 160* further from 160* take half to make inclined angle 80* to D and C. Now since two lines are parallel A intersects and D gets split to 40* each. so I got my final answer as 60* after using the Triangle = 180* ### You'll have to be more

You'll have to be more specific. You're using the fact that a central angle is twice the measure of an inscribed angle holding the same chord. However, I can't tell which angles you feel are 80 degrees.

Please specify some angles so I can better help.

### Hey Brent!

Hey Brent!
What do you think of this method?
AOD=180-DOB=160 degrees
Since OA=OD then OAD is an isosceles triangle and the angles at the base are equal
And since the triangle ADC is inscribed in the circle such that one of his sides is a diameter of the circle, it is a right triangle and ECD=90 degrees
We now have 2 out of 3 angles in the triangle containing the angle x
x=180-(10+90)=80 degrees ### That's a perfectly-reasoned

That's a perfectly-reasoned solution. Nice work!!

### For this question, AO = DO,

For this question, AO = DO, since the top angle is 160 and it is isos triangle two sides must be 10, so x = 180-90-10=80 Excellent!