In triangle ABC , let D be the mid point of BC. If ∠ADB = 45° and ∠ACD = 30° , determine ∠BAD.
Draw BL perpendicular to AC and join L to D. Since ∠BCL = 30° , we get ∠CBL = 60°.
Since BLC is a right triangle with ∠BCL = 30° , we have BL = BC /2 = BD. Thus in triangle BLD , we observe that BL = BD and ∠DBL= 60°.
This implies that BLD is an equilateral triangle hence LB = LD. Using ∠LDB = 60° and ∠ADB=45° , we get ∠ADL = 15°. But ∠DAL = 15° , thus LD = LA. Hence we have LD = LA = LB.
This implies that L is the circumcenter of the triangle BDA , thus ∠BAD = ∠BLD / 2 = 60° / 2 = 30°
Draw BL perpendicular to AC and join L to D. Since ∠BCL = 30° , we get ∠CBL = 60°.
Since BLC is a right triangle with ∠BCL = 30° , we have BL = BC /2 = BD. Thus in triangle BLD , we observe that BL = BD and ∠DBL= 60°.
This implies that BLD is an equilateral triangle hence LB = LD. Using ∠LDB = 60° and ∠ADB=45° , we get ∠ADL = 15°. But ∠DAL = 15° , thus LD = LA. Hence we have LD = LA = LB.
This implies that L is the circumcenter of the triangle BDA , thus ∠BAD = ∠BLD / 2 = 60° / 2 = 30°
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