Need some help!

[tawny mica]

Note that $DBC \sim ABC$ by $$\angle BAC= 90^{\circ}-\angle ABC = 90^{\circ}-(90^{\circ}-\angle DBC)=\angle DBC$$ Ratios give $$\frac{BC}{DC}=\frac{AC}{BC}\iff BC^2=AC\cdot DC$$ So from this we get $$BC^2=12\cdot 4 =48 \iff BC=4\sqrt{3}$$ So \boxed{$$BC = 4\sqrt{3}$$}

[sinful arch]

Wait but its ABC ~ ABD @tawny mica

Not DBC

[tawny mica]

Oh yeah I mixed up some points

Wait a moment

[sinful arch]

👍

[river saffron]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[polar turtle]

!nosols

[river saffron]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[daring rose]

Homelama

[tawny mica]

This should be fine @sinful arch

[polar turtle]

isn't that still a solution

[sinful arch]

I really thought this question would be treated as a binomial question where you stack 2

A^2 + B^2 = C^2

Hmmmm

[tawny mica]

Yeah this also works

[sinful arch]

Oh fr?

I remember the binomial method but im not sure about the ratio method

Would they or SHOULD they result in the same answer

[tawny mica]

Hmm wait let me think through this real quick

[sinful arch]

Im probably going to try using binomial stacking

Ok

[tawny mica]

Yeah you basically can apply Pythagorean theorem 3 times and you get a 3 equations with 3 unknowns, it’s just like more tedious than just ratios

But will result in the same thing

[sinful arch]

Oh ok

Let me try that

[sinful arch]

So this should be right though, is the “note that” a typo

@tawny mica

[tawny mica]

Yeah you can ignore that

[sinful arch]

So its a typo

Got it

👍

[solar comet]

.close