In the circle below , mA= 86, mBDC= 32, mAD= 48 find the mBC, mCD

Question

In the circle below, m∠A=86, m∠BDC=32, and mA͡D= 48 find mB͡C, mC͡D, mA͡B, m∠ADB, m∠ABD, m∠DBC, m∠BCD

Answer 1

Hints:

$mA͡D = 2m\angle ABD \implies m\angle ABD = {1\over2}mA͡D = \_\_$ by central angle theorem.

$mB͡C = 2m\angle BDC = \_\_$ by central angle theorem.

$\square ABCD$ is a cyclic quadrilateral $\implies m\angle BAD + m\angle BCD = 180^\circ \implies m\angle BCD = \_\_$

In $\triangle ABD, m\angle ABD + m\angle BAD + m\angle ADB = 180^\circ \implies m\angle ADB = \_\_$

In $\triangle DBC, m\angle DBC + m\angle DCB + m\angle BDC = 180^\circ \implies m\angle DBC = \_\_$

$mC͡D = 2m\angle CBD = \_\_$ by central angle theorem.

$m\angle ADB = \_\_ \implies mA͡B = \_\_$ by central angle theorem.