Trangles - Proof

In a triangle \( ABC \) , \( \angle A \) is twice of \( \angle B \) and \( a, b \) and \( c \) are respective sides opposite to angles \( \angle A, \angle B \) and \( \angle C \). then prove that \( a^2 = b(b+c) \)

#Geometry #Proofs #Triangles

Note by Muzaffar Ahmed
6 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Let D be on BC such that AD bisects ∠A\angle A. Then △ABD\triangle ABD is isoceles. Also, △ADC\triangle ADC is similar to △BAC\triangle BAC as all their corresponding interior angles are equal.

By angle bisector theorem CD=abb+c\frac {ab}{b+c}. However, since △ACD\triangle ACD is similar to △BCA\triangle BCA, ab=babb+c=b+ca\frac {a}{b}=\frac {b}{\frac {ab}{b+c}}=\frac {b+c}{a}. Cross multiplying gives the result.

Joel Tan - 6 years, 8 months ago

One of the proofs is this.. Does anyone have a simpler proof?

Let ∠A=2x \angle A = 2x , ∠B=x \angle B = x

From Sine rule,

SinAa=SinBb \frac{Sin A}{a} = \frac{Sin B}{b}

  ⟹  SinASinB=ab \implies \frac{Sin A}{Sin B} = \frac{a}{b}

  ⟹  Sin2xSinx=ab \implies \frac{Sin 2x}{Sin x} = \frac{a}{b}

  ⟹  2CosxSinxSinx=ab \implies \frac{2 Cos x Sin x}{Sin x} = \frac{a}{b}

  ⟹  2Cosx=ab \implies 2 Cos x = \frac{a}{b} ...(1)

From Cosine rule,

CosA=b2+c2−a22bc Cos A = \frac{b^2 + c^2 - a^2}{2bc} ...(2)

CosB=Cosx=a2+c2−b22ac Cos B = Cos x = \frac{a^2 + c^2 - b^2}{2ac} ...(3)

Now, CosA=Cos2x=Cos2x−Sin2x Cos A = Cos 2x = Cos^{2} x - Sin^{2} x

  ⟹  (Cos2x)+1 \implies (Cos 2x) + 1

=[Cos2x−Sin2]+[Cos2x+Sin2x]=2Cos2x = [ Cos^{2} x - Sin^{2} ] + [ Cos^{2} x + Sin^{2} x ] = 2 Cos^{2} x

From (1), 2Cos2x=(Cos2x)+1=b2+c2−a22bc+1 2 Cos^{2} x = ( Cos 2x ) + 1 = \frac{b^2 + c^2 - a^2}{2bc} + 1

=2Cos2x=b2+c2−a2+2bc2bc = 2 Cos^{2} x = \frac{b^2 + c^2 - a^2 + 2bc}{2bc} ...(4)

Dividing (4) with (3),

2Cos2xCosx \frac{2 Cos^{2} x}{Cos x}

=2Cosx=b2+c2−a2+2bca2+c2−b2×2ac2bc = 2 Cos x = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{2ac}{2bc}

=2Cosx=b2+c2−a2+2bca2+c2−b2×ab = 2 Cos x = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{a}{b}

From (1), 2Cosx=ab 2 Cos x = \frac{a}{b}

  ⟹  ab=b2+c2−a2+2bca2+c2−b2×ab \implies \frac{a}{b} = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{a}{b}

  ⟹  b2+c2−a2+2bca2+c2−b2=1 \implies \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} = 1

  ⟹  a2+c2−b2=b2+c2−a2+2bc \implies a^2 + c^2 - b^2 = b^2 + c^2 - a^2 + 2bc

  ⟹  2a2=2b2+2bc \implies 2a^2 = 2b^2 + 2bc

  ⟹  a2=b2+bc \implies a^2 = b^2 + bc

=a2=b(b+c) = \boxed{a^2 = b(b+c)}

Muzaffar Ahmed - 6 years, 8 months ago

Log in to reply

Nice use of trigonometry, although similar triangles would quicken the process

Joel Tan - 6 years, 8 months ago

zaffar u are super

Karthik Akondi - 6 years, 8 months ago
×

Problem Loading...

Note Loading...

Set Loading...