# 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 $$a^2 = b(b+c)$$

Note by Muzaffar Ahmed
3 years, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

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

- 3 years, 8 months ago

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

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

From Sine rule,

$$\frac{Sin A}{a} = \frac{Sin B}{b}$$

$$\implies \frac{Sin A}{Sin B} = \frac{a}{b}$$

$$\implies \frac{Sin 2x}{Sin x} = \frac{a}{b}$$

$$\implies \frac{2 Cos x Sin x}{Sin x} = \frac{a}{b}$$

$$\implies 2 Cos x = \frac{a}{b}$$ ...(1)

From Cosine rule,

$$Cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ ...(2)

$$Cos B = Cos x = \frac{a^2 + c^2 - b^2}{2ac}$$ ...(3)

Now, $$Cos A = Cos 2x = Cos^{2} x - Sin^{2} x$$

$$\implies (Cos 2x) + 1$$

$$= [ Cos^{2} x - Sin^{2} ] + [ Cos^{2} x + Sin^{2} x ] = 2 Cos^{2} x$$

From (1), $$2 Cos^{2} x = ( Cos 2x ) + 1 = \frac{b^2 + c^2 - a^2}{2bc} + 1$$

$$= 2 Cos^{2} x = \frac{b^2 + c^2 - a^2 + 2bc}{2bc}$$ ...(4)

Dividing (4) with (3),

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

$$= 2 Cos x = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{2ac}{2bc}$$

$$= 2 Cos x = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{a}{b}$$

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

$$\implies \frac{a}{b} = \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} \times \frac{a}{b}$$

$$\implies \frac{b^2 + c^2 - a^2 + 2bc}{a^2 + c^2 - b^2} = 1$$

$$\implies a^2 + c^2 - b^2 = b^2 + c^2 - a^2 + 2bc$$

$$\implies 2a^2 = 2b^2 + 2bc$$

$$\implies a^2 = b^2 + bc$$

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

- 3 years, 8 months ago

zaffar u are super

- 3 years, 8 months ago

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

- 3 years, 8 months ago