Question

In the diagram below of AACD, DB is a median to AC, and AB = DB.АBCIf mZDAB = 32°, what is mZBDC? Explain

Answer 1

Since the median divides the triangle ACD in two and they share an edge (the edge BD) we can use this information to try and find the angle mFor the triangle ABD we know that

$\frac{\sin m\angle DAB}{BD}=\frac{\sin m\angle ADB}{AB}$

Now, since we know that

$AB\cong DB$

then the last equation simplifies to

$\sin m\angle DAB=\sin m\angle ADB$

therefore m$\begin{gathered} m\angle DAB=32 \\ m\angle ADB=32 \\ m\angle ABD=116 \end{gathered}$For the triangle BDC we know that

$\frac{\sin m\angle BDC}{BC}=\frac{\sin m\angle BCD}{BD}$

but since the segment DB is the median, this means that AB is equal to BC, furthermore we know that AB is equal to DB, this means that

$BC\cong BD$

so, the last law of sines simplifies to

$\sin m\angle BDC=\sin m\angle BCD$

That implies that

$m\angle BDC=m\angle BCD$

Now we have to notice that the angle m$m\angle BCD=64$Finally we can conclude that the angle we are looking for is 32°.