Question

QUESTION IN ATTACHMENT!

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Answer 1

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

Given that,

~ E is the midpoint of BC

$\bf\implies \:BE = EC = \dfrac{1}{2}BC - - - (1)$

Now, In right triangle ABD

Using Pythagoras Theorem, we have

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

can be rewritten as

$\rm :\longmapsto\: {AB}^{2} = {AD}^{2} + {(BE + ED)}^{2}$

$\rm :\longmapsto\: {AB}^{2} = {AD}^{2} + {BE}^{2} + {ED}^{2} + 2(BE)(ED)$

can be re-arranged as

$\rm :\longmapsto\: {AB}^{2} = {AD}^{2} + {ED}^{2} + {BE}^{2} + 2(BE)(ED)$

$\bf :\longmapsto\: {AB}^{2} = {AE}^{2} + {BE}^{2} + BC \times ED - - - (2)$

$\red{\bigg \{ \because \:BE = \dfrac{1}{2}BC \bigg \}}$

and

$\red{\bigg \{ \because \: {AD}^{2} + {ED}^{2} = {AE}^{2} \bigg \}}$

Now, In right triangle ACD

By using Pythagoras Theorem, we have

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

can be rewritten as

$\rm :\longmapsto\: {AC}^{2} = {(CE - ED)}^{2} + {AD}^{2}$

$\rm :\longmapsto\: {AC}^{2} = {CE}^{2} + {ED}^{2} - 2(CE)(ED) + {AD}^{2}$

can be re-arranged as

$\rm :\longmapsto\: {AC}^{2} = {AD}^{2} + {ED}^{2} - 2(CE)(ED) + {CE}^{2}$

$\bf :\longmapsto\: {AC}^{2} = {AE}^{2} -BC \times ED + {CE}^{2} - - - (3)$

On Subtracting equation (3) from equation (2), we get

$\rm :\longmapsto\:{AB}^{2}- {AC}^{2} = \cancel{AE}^{2} + BC \times ED + {BE}^{2} -\cancel{AE}^{2} + BC \times ED - {CE}^{2}$

$\rm :\longmapsto\:{AB}^{2}- {AC}^{2} = 2BC \times ED + \cancel{BE}^{2} -\cancel {BE}^{2}$

$\red{\bigg \{ \because \:BE = EC \bigg \}}$

Hence,

$\boxed{ \bf{ \: {AB}^{2} - {AC}^{2} = 2BC \times ED}}$

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,

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.