Question

in triangle abc ad is perpendicular bisector of BC and BD equal to one third of CD prove that 2CA square equals to a b square + BC square ​

Answer 1

ANSWER:-

Given:

In the triangle, AD⊥BC & BD= 1/3CD.

To prove:

Prove that 2CA²= 2AB² +BC²

Proof:

We have,

BC= BD + CD

BC= 1/3CD + CD

BC= 1+3CD/3

BC= 4CD/3 [as BD= 1/3CD]

=) CD= 3/4BC........(1)

As,

AD⊥BC

=) ∆ADC is a right-angled triangle.

By using Pythagoras Theorem;

(hypotenuse)²=(base)² +(perpendicular)²

=) AD² + CD² = CA²

=) AD² = CA² - CD².......(2)

Therefore,

∆ABD is a right angled triangle.

By using Pythagoras Theorem:

AD² +BD² = AB²

CA² -CD² + BD² = AB² ......[From (2)]

$= > {CA}^{2} - {CD}^{2} + ( \frac{1}{3} CD) {}^{2} = {AB}^{2} \: \: \: \: (As \: BD \ = \frac{1}{3} CD) \\ \\ = > {CA}^{2} - {CD}^{2} + \frac{1}{9} {CD}^{2} = {AB}^{2} \\ \\ = > {CA}^{2} - \frac{8}{9} {CD}^{2} = {AB}^{2} \\ \\ = > {CA}^{2} - \frac{8}{9} ( \frac{3}{4} BC) {}^{2} = {AB}^{2} \: \: \: \: [from \: (1)] \\ \\ = > {CA}^{2} - \frac{8}{9} \times \frac{9}{16} \times {BC}^{2} = {AB}^{2} \\ \\ = > {CA}^{2} - \frac{1}{2} {BC}^{2} = {AB}^{2} \\ \\ = > 2 {CA}^{2} - {BC}^{2} = 2 {AB}^{2} \\ \\ = > 2{CA}^{2} = 2 {AB}^{2} + {BC}^{2}$

Hence,

Proved.

Hope it helps ☺️