Question

Prove: In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.'​

Answer 1

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

Let assume a right-angled triangle ABC, right-angled at A.

Let AD be the perpendicular drawn from vertex A on hypotenuse BC, intersecting BC at D.

We have to prove that $\bf \:{AD}^{2}=BD\:\times\:CD$

Now, In right-angle triangle ABC,

By using Pythagoras Theorem, we have

$\rm \: {AB}^{2} + {AC}^{2} = {BC}^{2} - - - (1)$

Now, In right-angle triangle ABD

Using Pythagoras Theorem, we have

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

Now, In right-angle triangle ACD,

Using Pythagoras Theorem, we have

$\rm \: {AC}^{2} = {AD}^{2} + {CD}^{2} - - - (3)$

On adding equation (2) and (3), we get

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

Now, using equation (1), the above equation can be rewritten as

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

can be further rewritten as

$\rm \: {(BD + CD)}^{2} = 2 {AD}^{2} + {BD}^{2} + {CD}^{2}$

$\rm \: {BD}^{2} + {CD}^{2} + 2(BD)(CD) = 2 {AD}^{2} + {BD}^{2} + {CD}^{2}$

$\rm \: 2(BD)(CD) = 2 {AD}^{2}$

$\bf\implies \: (BD) \times (CD) = {AD}^{2}$

$\bf\implies \:AD \: is \: geometric \: mean \: between \: BD \: and \: CD.$

Hence,

In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

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

Additional Information :-

1. $\rm\:Pythagoras \:Theorem$ :-

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

Answer:

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

Let assume a right-angled triangle ABC, right-angled at A.

Let AD be the perpendicular drawn from vertex A on hypotenuse BC, intersecting BC at D.

We have to prove that $\bf \:{AD}^{2}=BD\:\times\:CD$

Now, In right-angle triangle ABC,

By using Pythagoras Theorem, we have

$\rm \: {AB}^{2} + {AC}^{2} = {BC}^{2} - - - (1)$

Now, In right-angle triangle ABD

Using Pythagoras Theorem, we have

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

Now, In right-angle triangle ACD,

Using Pythagoras Theorem, we have

$\rm \: {AC}^{2} = {AD}^{2} + {CD}^{2} - - - (3)$

On adding equation (2) and (3), we get

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

Now, using equation (1), the above equation can be rewritten as

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

can be further rewritten as

$\rm \: {(BD + CD)}^{2} = 2 {AD}^{2} + {BD}^{2} + {CD}^{2}$

$\rm \: {BD}^{2} + {CD}^{2} + 2(BD)(CD) = 2 {AD}^{2} + {BD}^{2} + {CD}^{2}$

$\rm \: 2(BD)(CD) = 2 {AD}^{2}$

$\bf\implies \: (BD) \times (CD) = {AD}^{2}$

$\bf\implies \:AD \: is \: geometric \: mean \: between \: BD \: and \: CD.$

Hence,

In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

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

Additional Information :-

1. $\rm\:Pythagoras \:Theorem$ :-

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.