In Fig., ABC is a triangle in which ∠B = 2∠C. D is a point on side

1.	In Fig., ABC is a triangle in which ∠B = 2∠C. D is a point on side such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is[Hint: Δ ABE ≅ Δ DCE]A. 72°B. 73°C. 74°D. 95°
Answer» Given that ΔABC BE is bisector of ∠Band AD is bisector of ∠BAC ∠B = 2 ∠C By exterior angle theorem in triangle ADC ∠ADB = ∠DAC + ∠C (i) In ΔADB, ∠ABD + ∠BAD + ∠ADB = 180° 2 ∠C + ∠BAD + ∠DAC + ∠C = 180°[From (i)] 3 ∠C + ∠BAC = 180° ∠BAC = 180° – 3 ∠C (ii) Therefore, AB = CD ∠C = ∠DAC ∠C = \(\frac{1}{2}\) ∠BAC (iii) Putting value of Angle C in (ii), we get ∠BAC = 180° – \(\frac{1}{2}\) ∠BAC ∠BAC + \(\frac{3}{2}\)∠BAC = 180° \(\frac{5}{2}\)∠BAC = 180° ∠BAC = \(\frac{180\times 2}{5}\) = 72° ∠BAC = 72°

In Fig., ABC is a triangle in which ∠B = 2∠C. D is a point on side such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is[Hint: Δ ABE ≅ Δ DCE]A. 72°B. 73°C. 74°D. 95°

Answer»

Given that ΔABC

BE is bisector of ∠Band AD is bisector of ∠BAC

∠B = 2 ∠C

By exterior angle theorem in triangle ADC

∠ADB = ∠DAC + ∠C (i)

In ΔADB,

∠ABD + ∠BAD + ∠ADB = 180°

2 ∠C + ∠BAD + ∠DAC + ∠C = 180°[From (i)]

3 ∠C + ∠BAC = 180°

∠BAC = 180° – 3 ∠C (ii)

Therefore,

AB = CD

∠C = ∠DAC

∠C = \(\frac{1}{2}\) ∠BAC (iii)

Putting value of Angle C in (ii), we get

∠BAC = 180° – \(\frac{1}{2}\) ∠BAC

∠BAC + \(\frac{3}{2}\)∠BAC = 180°

\(\frac{5}{2}\)∠BAC = 180°

∠BAC = \(\frac{180\times 2}{5}\)

= 72°

∠BAC = 72°