Question

In Fig. 10.141, 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 1

∠BAC = 72°

Step-by-step explanation:

∠B = 2∠C

BE is the bisector of ∠B

=> ∠CBE = ∠B/2 = 2∠C/2 = ∠C

=> BE = EC

∠AEB = ∠CBE + ∠C = 2∠C

in Δ ABE

=> ∠ABE + ∠AEB   + ∠A = 180°

=> ∠C + 2∠C +  ∠A = 180°

=> ∠A = 180° - 3∠C

From given options only  72° will satisfy this

Hence ∠BAC = 72°

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Answer 2

Answer:

ZBAC = 72

Step-by-step explanation:

ZB=22C

BE is the bisector of ZB

=> ZCBE = ZB/2 = 2/C/2 = ZC

=> BE = EC

ZAEB = ZCBE + ZC = 24C

in A ABE

=> ZABE+ZAEB + ZA = 180°

=> ZC+2/C+ ZA = 180°

=> ZA = 180° - 3/C

From given options only 72° will satisfy this

Hence ZBAC = 72°