Question:
In figure 9.35, AB divides ∠DAC in the ratio 1: 3 and AB = DB. Determine the value of x.
Solution:
Let ∠BAD = Z, ∠BAC = 3Z
⇒ ∠BDA = ∠BAD = Z (∵ AB = DB)
Now ∠BAD + ∠BAC + 108° = 180° [Linear pair]
⇒ Z + 3Z + 108° =180°
⇒ 4Z = 72°
⇒ Z = 18°
Now, In ΔADC
∠ADC + ∠ACD = 108° [Exterior angle property]
⇒ x + 18° = 180°
⇒ x = 90°
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