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ABC is a triangle in which ∠B = 2 ∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°. - Mathematics

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प्रश्न

ABC is a triangle in which ∠B = 2 ∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD.
Prove that ∠BAC = 72°. 

योग
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उत्तर

In ΔABC,
∠B = 2∠C or, ∠B = 2y, where ∠C = y.

AD is the bisector of ∠BAC.
So, let ∠BAD = ∠CAD = x.

Let BP be the bisector of ∠ABC. Join PD.

In ΔBPC, we have
∠CBP = ∠BCP = y ⇒ BP = PC

In Δ′s ABP and DCP, we have
∠ABP = ∠DCP, 
∠ABP = ∠DCP = y

AB = DC     ........(Given) and,
BP = PC      ........(As proved above)
So, by SAS congruence criterion, we obtain
ΔABP ≅ ΔDCP
⇒ ∠BAP = ∠CDP and AP = DP
⇒ ∠CDP = 2x and ∠ADP = DAP = x ......[∴∠A = 2x]
In ΔABD, we have
∠ADC = ∠ABD + ∠BAD
⇒ x + 2x = 2y + x
⇒ x = y
In ΔABC, we have
∠A + ∠B + ∠C = 180
⇒ 2x + 2y + y = 180
⇒ 5x = 180    ......[∵ x = y]
⇒ x = 36
Hence, ∠BAC = 2x = 72.
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अध्याय 12: Congruent Triangles - Exercise 12.3 [पृष्ठ ४७]

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आरडी शर्मा Mathematics [English] Class 9
अध्याय 12 Congruent Triangles
Exercise 12.3 | Q 10 | पृष्ठ ४७

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