In ∆ABC the angle bisectors drawn from vertices A and B intersect at point D. Find m∠ADB if: m∠C=γ

  ∠ADB = γ/2 +90°

Step-by-step explanation:

Here's one way to show the measure of ∠ADB.

  ∠ADB = 180° - (α + β) . . . . . sum of angles in ΔABD

  ∠ADB + (2α +β) + γ + (2β +α) = 360° . . . . . sum of angles in DXCY

Substituting for (α + β) in the second equation, we get ...

  ∠ADB + 3(180° - ∠ADB) + γ = 360°

  180° + γ = 2(∠ADB) . . . . . . add 2(∠ADB)-360°

  ∠ADB = γ/2 + 90° . . . . . . . divide by 2

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To find angles CXD and CYD, we observe that these are exterior angles to triangles AXB and AYB, respectively. As such, those angles are equal to the sum of the remote interior angles, taking into account that AY and BX are angle bisectors.

y=mx+b is the final answer!   : )

what are ur choices?