In ∆abc the angle bisectors drawn from vertexes a and b intersect at point d. find ∠adb if: d m∠c = γ m∠adb =

In ∆ABD, ∠BAD = (∠BAC)/2 and ∠ABD = (∠ABC)/2.

The sum of angles in that triangle is 180°, so

... ∠ADB + ∠BAD + ∠ABD = 180°

Substituting the above relations, we have ...

... ∠ADB + (∠BAC)/2 + (∠ABC)/2 = 180°

... ∠ADB = 180° - (∠BAC + ∠ABC)/2

Turning our attention now to ∆ABC, its angles, too, add to 180°.

... ∠BAC + ∠ABC + γ = 180°

We can subtract γ to get

... ∠BAC + ∠ABC = 180° - γ

Substituting this expression into the above expression for ∠ADB, we have

... ∠ADB = 180° - (180° - γ)/2

... ∠ADB = 90° + γ/2

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