Prove that the diagonals of a rhombus are also angle bisectors. (theorem 1)

A rhombus has four congruent sides.  Without loss of generality, we denote the rhombus by the consecutive points PQRS.

1. Draw diagonal PR.
2. Consider triangle PQR.  We conclude that PQ=RQ because a rhombus has four congruent sides.  Triangle PRQ is isosceles => angles QPR and QRP are congruent.
3. Similarly, PS=RS. Triangle PSR is isosceles.
4. PR (diagonal) is common side, so triangles PQR is congruent to PSR (SSS).
5. Therefore angles QRP=QPR=SPR=SRP.
6. Since angles QPR=SPR, and SPR=QPR, we conclude that the diagonal PR is an angle bisector.

Finally, using similar arguments, we can show that diagonal QS bisects angles at Q and S.

It is 198,87032,750 + 375 = 33145 so, we take the 6% and we make it a decimal (0.6) and multiply that, then that is your answer