Question
Find the value of ∠ABD if AB = AC and DB = DC
- 20°
- 45°
- 30°
- 15°
Hint:
From figure, in ΔDBC, DB = DC ⇒ ∠DBC = ∠DCB (Angles opposite to equal sides are equal)
Also, ∠DBC + ∠DCB + ∠BDC = 180° (Angle sum property of triangle)
In ΔABC, AB = AC ⇒ ∠ABC = ∠ACB (Angles opposite to equal sides are equal)
Also, ∠ABC + ∠ACB + ∠BAC = 180° (Angle sum property of triangle)
∠ABD = ∠ABC – ∠DBC
The correct answer is: 15°
From figure, in ΔDBC, DB = DC ⇒ ∠DBC = ∠DCB (Angles opposite to equal sides are equal)
Also, ∠DBC + ∠DCB + ∠BDC = 180° (Angle sum property of triangle)
⇒ 80° + 2∠DBC = 180°
⇒ ∠DBC = 50° ————– (i)
In ΔABC, AB = AC ⇒ ∠ABC = ∠ACB (Angles opposite to equal sides are equal)
Also, ∠ABC + ∠ACB + ∠BAC = 180° (Angle sum property of triangle)
⇒ 50° + 2∠ABC = 180°
⇒ ∠ABC = 65° ————– (i)
Now, ∠ABD = ∠ABC – ∠DBC = 65° – 50° = 15°.
Also, ∠DBC + ∠DCB + ∠BDC = 180° (Angle sum property of triangle)
⇒ 80° + 2∠DBC = 180°
⇒ ∠DBC = 50° ————– (i)
In ΔABC, AB = AC ⇒ ∠ABC = ∠ACB (Angles opposite to equal sides are equal)
Also, ∠ABC + ∠ACB + ∠BAC = 180° (Angle sum property of triangle)
⇒ 50° + 2∠ABC = 180°
⇒ ∠ABC = 65° ————– (i)
Now, ∠ABD = ∠ABC – ∠DBC = 65° – 50° = 15°.
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