ABC is a triangle in which angle B =2angle C. D is a point on BC such that AD is the angle bisector of angleA and AB=CD.Prove that angle A is 72°
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Answers
Step-by-step explanation:
In ΔABC, we have
∠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, we have
∠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∘
Answer:
Let us consider a positive integer a
Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that
a = 3b + r……………………………(1)
where r = 0,1,2,3…..
Case 1: Consider r = 0
Equation (1) becomes
a = 3b
On squaring both the side
a2 = (3b)2
a2 = 9b2
a2 = 3 × 3b2
a2 = 3m
Where m = 3b2
Case 2: Let r = 1
Equation (1) becomes
a = 3b + 1
Squaring on both the side we get
a2 = (3b + 1)2
a2 = (3b)2 + 1 + 2 × (3b) × 1
a2 = 9b2 + 6b + 1
a2 = 3(3b2 + 2b) + 1
a2 = 3m + 1
Where m = 3b2 + 2b
Case 3: Let r = 2
Equation (1) becomes
a = 3b + 2
Squaring on both the sides we get
a2 = (3b + 2)2
a2 = 9b2 + 4 + (2 × 3b × 2)
a2 = 9b2 + 12b + 3 + 1
a2 = 3(3b2 + 4b + 1) + 1
a2 = 3m + 1
where m = 3b2 + 4b + 1
∴ square of any positive integer is of the form 3m or 3m+1.
Hence proved.