Question

(ii) In A ABC, seg AD perpendicular seg BC. Prove : AB square+CDsquare=BDsquare+ ACsquare
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Answer 1

Question :- In triangle ABC, segment AD is perpendicular to BC. Prove that

$\rm \: {AB}^{2} + {CD}^{2} = {BD}^{2} + {AC}^{2}$

$\large\underline{\sf{Solution-}}$

As it is given that

$\rm \: AD \: \perp \: BC$

$\rm\implies \angle\:ADB \: = \: \angle ADC \: = \: 90 \degree$

Now, in right angle triangle ADB

By using Pythagoras Theorem, we have

$\rm \: {AB}^{2} = {AD}^{2} + {BD}^{2}$

$\rm\implies \: {AD}^{2} = {AB}^{2} - {BD}^{2} - - - (1)$

Now, in right angle triangle ADC

By using Pythagoras Theorem, we have

$\rm \: {AC}^{2} = {AD}^{2} + {CD}^{2}$

$\rm\implies \: {AD}^{2} = {AC}^{2} - {CD}^{2} - - - (2)$

From equation (1) and (2), we get

$\rm \: {AB}^{2} - {BD}^{2} = {AC}^{2} - {CD}^{2}$

$\rm\implies \:\boxed{ \rm{ \: {AB}^{2} + {CD}^{2} = {BD}^{2} + {AC}^{2} \: \: }}$

Hence, Proved

$\rule{190pt}{2pt}$

Additional Information :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem,

This theorem states that : - If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

by cancelling AD from right hand side we get BD square -CD square

and interchanging the position of negative segments we can prace this proof