How do I do this?

[trim terrace]

What is the trick to find AD? <@&286206848099549185>

[viscid stream]

you could draw line CD which will be equal in length to AC because it is also a radius, and so will have length 3

and then use trigonometry to find angle CAD

which will be the same as angle ADC

becuase triangle ADC is isoceles

and then from there you can find the last angle ACD

and use law of sines

or cosine

to find AD

[trim terrace]

my teacher says not to use trigonometry

[viscid stream]

oh

it is a 3 4 5 right triangle

making angle CAD 60 degrees

[trim terrace]

no, 3-4-5 doesn't mean 30-60-90 triangle

[viscid stream]

no?

[viscid stream]

oh right

another way...

[trim terrace]

<@&286206848099549185>

[versed holly]

hey

@trim terrace saw the problem

so ABC is a 3-4-5 triangle; do you understand that from pythagorean theorem?

[trim terrace]

yes AB = 5 by pythagorean triple

[versed holly]

alright

now lets focus on the line segment CB

now we know that CB is 4 units long; correct?

[trim terrace]

give me one second, brb

ok yes i am back

yes

[versed holly]

actually hold on a minute

okay actually first I want you to draw an extra line from C to D

notice anything about the triangle ACD?

[trim terrace]

acd is not 90-45-45 because angle C is getting cut

so doesn't work...

[versed holly]

okay, that does not matter

What do you notice is special about ACD in general?

[trim terrace]

yes isosceles

whats next

[versed holly]

Right

now, what if you drew the perpendicular bisector from C to AD?

[trim terrace]

i don't know what u are hinting at

[versed holly]

okay. so lets drwa the perpendicular bisector, and call the point where it hits AD point F.

triangle ACF is a right triangle. Correct?

[trim terrace]

yes

[versed holly]

okay. now do you see anything in common with triangle ACF and triangle ABC?

hint: angles

[trim terrace]

yes similarity

continue

[versed holly]

right. which means that ACF is ALSO a 3-4-5 triangle. correct?

[trim terrace]

huhhhhh

those 2 are similar not congruent

[versed holly]

okay, well 3-4-5 as in side ratios, not lengths

[trim terrace]

yes

[versed holly]

okay, now taking into account that triangle ACF is a 3-4-5 right triangle with hypotenuse 3 (remember, AC is 3), how would you find AF?

[trim terrace]

ok 1st of all why would AF be needed we want AD

[versed holly]

well remember how AF is A connected to the point where the perpendicular biseictor of AD meets the line?

[trim terrace]

ok i see, can you help me with another question now?

[versed holly]

sure. in conclusion, we used the perpendicular bisector of AD to split it up into two equal parts, then used similar triangles to find one of those halves and then AD as a whole.

[viscid stream]

how do you know that the perpendicular bisector of AD will intersect the point C

[trim terrace]

start from point C and draw perpendicular line to AB

[versed holly]

because ACD is an isoceles triangle

[viscid stream]

oh its an altitude right?

[trim terrace]

and exploiting the circle equal radius property

[viscid stream]

right?

[trim terrace]

that doesn't necessary mean it gonna be bisector, its bisector because of isosceles triangle

[viscid stream]

wdym

wait

hm?

[versed holly]

yes

[versed holly]

so if you have an isoceles triangle, the two legs are equal, rendering it symmetrical

therefore, the median of the triangle down the legs also happpens to be te perpendicular bisector

that is why it intersects c

[viscid stream]

oh