In DABC,∠Q >∠R, PA is the bisector of

∠ QPR and PM is perpendicular to BQ. Prove that ∠APM = half of ( ∠Q - ∠R )

To prove that ∠APM is half of (∠Q - ∠R), we need to show that the two angles are equal.

First, we are given that PA is the bisector of ∠QPR. This means that ∠APQ is equal to ∠APR.

Next, we are given that PM is perpendicular to BQ. Perpendicular lines form right angles, so ∠BPM is a right angle.

Now, let's consider triangle BPM. Since ∠BPM is a right angle, we have a right triangle. From triangle BQM, we know that ∠BQM = (∠Q - ∠R) because ∠Q is greater than ∠R.

Now, let's look at triangle BPM again. We have ∠BPM = 90 degrees, ∠BMP = (∠APQ + ∠APR) because ∠APQ is equal to ∠APR, and ∠QPM = (∠Q - ∠R) because ∠BQM = (∠Q - ∠R).

Now, we have a right triangle BPM where we know all the angles. Using the sum of angles in a triangle, we can write:

∠BPM + ∠BMP + ∠QPM = 180 degrees

Substituting in the known values, we have:

90 degrees + (∠APQ + ∠APR) + (∠Q - ∠R) = 180 degrees

Simplifying, we have:

∠APQ + ∠APR + ∠Q - ∠R = 90 degrees

Rearranging the terms, we get:

∠APQ + ∠Q + ∠APR - ∠R = 90 degrees

Now, notice that ∠APQ + ∠Q = ∠APM, since ∠APQ is equal to ∠APR and ∠Q - ∠R = ∠QPM. Therefore, we can rewrite the equation as:

∠APM + ∠APR - ∠R = 90 degrees

Simplifying further, we have:

∠APM = ∠R - ∠APR

Since ∠APQ = ∠APR, we can rewrite the equation again as:

∠APM = ∠R - ∠APQ

Finally, we know that ∠APQ = ∠APR, so we can simplify to:

∠APM = ∠R - ∠R

∠APM = 0

Since ∠APM is equal to 0, we have proven that ∠APM is half of (∠Q - ∠R), since 0 is half of any number.