Question

can anyone solve:
please

Answer 1

∠ABD = 68°,
So, ∠DBC = 180° - 68° = 112°

In ∆BDC, BD = BC (Given)
Thus, ∆BDC is an isosceles triangle
In Isosceles Triangle, 2 base angles will be equal.
So, ∠BDC = ∠BCD

Thus, in ∆BDC
∠DBC + ∠BDC + ∠BCD = 180°
112° + ∠BDC + ∠BCD = 180°
∠BDC + ∠BCD = 180° - 112°
∠BDC + ∠BCD = 68°
2∠BCD = 68°
∠BCD = ∠BDC = 68/2 = 34°

Since, BD bisects ∠ADC
Thus, ∠BDC = ∠ADB = 34°

In ∆ABD,
∠ADB + ∠ABD + ∠BAD = 180°
34° + 68° + ∠BAD = 180°
102° + ∠BAD = 180°
∠BAD = 180° - 102 ° = 78°

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