Question

In the diagram, $BP$ and $BQ$ trisect $\angle ABC$. $BM$ bisects $\angle PBQ$. Find the ratio of the measure of $\angle MBQ$ to the measure of $\angle ABQ$.

Answer 1

1:4

Explanation:

`Let $\angle MBQ = x$, so $\angle MBP=x$ as well. Therefore, we have $\angle PBQ = 2x$, so $\angle ABP = \angle PBQ = \angle QBC = 2x$. Finally, we have $\angle ABQ = \angle ABP + \angle PBQ = 4x$, so\[(\angle MBQ)/(\angle ABQ) = (x)/(4x) = \boxed{\frac14}.\]`

Hope this helped! :)

Answer 2

1:4

Explanation:

If BP and BQ trisect angle ABC, then

`m\angle ABP=m\angle PBQ=m\angle QBC`

If BM bisects the angle MBQ, then

`m\angle PBM=m\angle MBQ=x^(\circ)`

Now, find the measure of angle ABQ in terms of x. First, note that

`m\angle PBQ=2m\angle MBQ=2x^(\circ)`

Now,

`m\angle ABQ=m\angle ABP+m\angle PBQ=2m\angle PBQ=2\cdot 2x^(\circ)=4x^(\circ)`

So, the ratio of the measure of angle MBQ to the measure of angle ABQ is

`(x^(\circ))/(4x^(\circ))=(1)/(4)`