Given $AB = 2$ and $AC = 3$, what are $AD$, $BD$, $m∠BAD$ and $m∠ABD$? [closed]

Solution 1:

Firstly take a look at triangle $ABD$, we know that the length $AT$ is $3$. Now we see that the lengths $BD = TD$ and we can let these lengths be $x$ in the diagram below(I added new variables T,W,E)

Hence, we apply Pythagoras since:

$$2^2 + x^2 = (3-x)^2.$$ Since the lenght $BD = x$ and $AD = AT - DT = 3 - x$. Hence,

Expanding and simplifying it is easy to see that $x = \frac{5}{6}$ hence the length $AD = 3 - \frac{5}{6}$ and the length $BD = \frac{5}{6}$. Does this line of reasoning help you to find what 𝑚∠𝐵𝐴𝐷 and 𝑚∠𝐴𝐵 are? I personally do not see a way in which the $\sqrt{13}$ fits into this, though there may be a way.