**✩** You must already know

- The Pythagoras Theorem
- Area of Triangles
- Simple Algebraic Equations

## Questions

Try to use a minimum number of hints to reach the solution. All the best!

If you find these challenging, learn to solve the Pythagoras problems here.

### 1 Question

In the right triangle ABC, point D is equidistant from AB and BC and AD = 30, DC = 40. Find lengths of sides AB and BC.

#### Answer

Diagram and Variables

Let the distance of D from AB be x

∴ ED = DF = x (D is equidistant from AB BC)

also BF = ED = x and EB = DF = x

Let AE = z ∴ AB = AE + EB = z + x

Let FC = y ∴ BC = BF + FC = x + y

Equating the area of triangle with its parts can you find the relation between x, y and z?

### 2 Question

In above right triangle:

Angle ACB is 90 degrees and sides AC = m, BC = n and AB = p

Altitude CD = h

AD = m_{1} and DB = n_{1}

Prove that:

1. h^{2} = m_{1}.n_{1}

2. m^{2} = n_{1}.p

3. n^{2} = m_{1}.p

4. $h=pmn $

#### Answer

1. h^{2} = m_{1}.n_{1}

Can you find relation between h, m_{1}, m using △ ACD?

### 3 Question

Prove that in a right triangle, the hypotenuse is twice the median drawn to the hypotenuse.

#### Answer

Let ABC be the right triangle.

Side AB = a, BC = b.

BD = Median to hypotenuse, so AD = DC = m

Construct altitude BE. Let ED = x

Let BE = h, BD = y

So we have to prove y = m

Can you now form three equations using triangles BDE, BEA and BCE and link x, h, y with a, b and m?

### 4 Question

In a right triangle ABC, length of the medians to the sides AB and BC are $261 $ and $601 $ respectively. Find the length of its hypotenuse.

#### Answer

Given:

AD = $601 $

CE = $261 $

Let AE = EB = x

BD = DC = y

Can you use the Pythagorean Theorem to form an equation between x, y, and AD?

## Answers

### 1 Answer

Diagram and Variables

Let the distance of D from AB be x

∴ ED = DF = x (D is equidistant from AB BC)

also BF = ED = x and EB = DF = x

Let AE = z ∴ AB = AE + EB = z + x

Let FC = y ∴ BC = BF + FC = x + y

Equations

Area of △ AED + Area of square EBFD + Area of △ DFC = Area of △ ABC

$2zx +x_{2}+2xy =2(z+x)(x+y) $

zx + 2x^{2} + xy = (z + x)(x + y)

zx + 2x^{2} + xy = zx + zy + x^{2} + xy

2x^{2} = zy + x^{2}

x^{2} = zy (Equation 1)

Applying the Pythagoras theorem on △ AED :

z^{2} + x^{2} = AD^{2} = 30^{2} = 900

z^{2} + x^{2} = 900

Replacing x^{2}, using equation 1:

z^{2} + zy = 900

z(z + y) = 900 (Equation 2)

Applying the Pythagoras theorem on △ DFC :

y^{2} + x^{2} = DC^{2} = 40^{2} = 1600

y^{2} + x^{2} = 1600

Replacing x^{2} by zx, using equation 1:

y^{2} + zy = 1600

y(y + z) = 1600 (Equation 3)

Dividing equation 3 by 2

$z(y+z)y(y+z) =9001600 $

$zy =916 $

$y÷z=16÷9$

$y/z=16/9$

$y=916z $ (Equation 4)

$y=16z/9$ (Equation 4)

Solve for the sides of the right triangle.

Using this value of y in equation 2:

$z(z+916z )=900$

$z_{2}+916z_{2} =900$

$925z_{2} =900$

$25z_{2}=900×9$

$z_{2}=25900×9 $

$z=530×3 =6×3=18$

From equation 4:

$y=916×18 =16×2=32$

Using equation 1:

x^{2} = zy = 18 × 32 = 9 × 64

x = 3 × 8 = 24

Therefore side AB = z + x = 18 + 24 = 42

and BC = x + y = 24 + 32 = 56

### 2 Answer

1. h^{2} = m_{1}.n_{1}

From △ ACD :

h^{2} + m_{1}^{2} = m^{2}

h^{2} = m^{2} − m_{1}^{2} Equation 1

From △ CDB :

h^{2} + n_{1}^{2} = n^{2}

h^{2} = n^{2} − n_{1}^{2} (Equation 2)

Adding Equation 1 and 2:

h^{2} + h^{2} = m^{2} − m_{1}^{2} + n^{2} − n_{1}^{2}

h^{2} + h^{2} = m^{2} + n^{2} − m_{1}^{2} − n_{1}^{2}

2h^{2} = m^{2} + n^{2} − m_{1}^{2} − n_{1}^{2} (Equation 3)

From △ ABC:

m^{2} + n^{2} = AB^{2} = (m_{1} + n_{1})^{2} as AB = m_{1} + n_{1}

Using this result to replace m^{2} + n^{2} in Equation 3

2h^{2} = (m_{1} + n_{1})^{2} − m_{1}^{2} − n_{1}^{2}

2h^{2} = m_{1}^{2} + n_{1}^{2} + 2m_{1}n_{1} − m_{1}^{2} − n_{1}^{2}

2h^{2} = 2m_{1}n_{1}

h^{2} = m_{1}n_{1}

Proved!

2. m^{2} = n_{1}.p

From △ ACD:

m^{2} = h^{2} + m_{1}^{2}

Using ( h^{2} = m_{1}.n_{1}), replace h^{2}:

m^{2} = m_{1}n_{1} + m_{1}^{2}

m^{2} = m_{1}(n_{1} + m_{1})

Now m_1 + n_1 = AB = p so

m^{2} = m_{1}(p)

h^{2} = m_{1}.p

Proved!

3. n^{2} = m_{1}.p

If you have done the previous proof, you don’t need hints for this one! The proof is very similar.

4. $h=pmn $

Multiplying the results m^{2} = n_{1}p and n^{2} = m_{1}p:

m^{2}n^{2} = (n_{1}p)(m_{1}p)

m^{2}n^{2} = n_{1}m_{1}p^{2}

Replacing m_{1}n_{1} by h^{2}:

m^{2}n^{2} = h^{2}p^{2}

$p_{2}m_{2}n_{2} =h_{2}$

$pmn =h$

Proved!

Another very simple way to prove this is by using the area of the triangle. I leave it to you to try it out.

### 3 Answer

Let ABC be the right triangle.

Side AB = a, BC = b.

BD = Median to hypotenuse, so AD = DC = m

Construct altitude BE. Let ED = x

Let BE = h, BD = y

So we have to prove y = m

Equations

From triangle BDE

BD^{2} = BE^{2} + ED^{2}

y^{2} = h^{2} + x^{2} (Equation 1)

From triangle BEA:

AB^{2} = BE^{2} + AE^{2}

a^{2} = h^{2} + (m − x)^{2}

h^{2} = a^{2} − (m − x)^{2} (Equation 2)

Similarly form equation from triangle BCE:

BC^{2} = BE^{2} + EC^{2}

b^{2} = h^{2} + (m + x)^{2}

h^{2} = b^{2} − (m + x)^{2} (Equation 3)

Solve for y

Adding equation 2 and 3:

2h^{2} = a^{2} − (m − x)^{2} + b^{2} − (m + x)^{2}

2h^{2} = a^{2} + b^{2} − [(m − x)^{2} + (m + x)^{2}]

2h^{2} = a^{2} + b^{2} − [m^{2} + x^{2} − 2mx + m^{2} + x^{2} + 2mx]

2h^{2} = a^{2} + b^{2} − [2m^{2} + 2x^{2}]

2h^{2} = a^{2} + b^{2} − 2m^{2} − 2x^{2}

2h^{2} + 2x^{2} = a^{2} + b^{2} − 2m^{2}

Dividing both side by 2:

$h_{2}+x_{2}=2a_{2}+b_{2}−2m_{2} $

Using equation 1, we replace h^{2} + x^{2} by y^{2}

$y_{2}=2a_{2}+b_{2}−2m_{2} $ (Equation 4)

Now from right triangle ABC, we know that:

a^{2} + b^{2} = AC^{2} = (2m)^{2} (as AC =AD + DC = m+m = 2m)

a^{2} + b^{2} = 4m^{2} (Equation 5)

Putting value of a^{2} + b^{2} in equation 4:

$y_{2}=24m_{2}−2m_{2} $

$y_{2}=22m_{2} $

y^{2} = m^{2}

y = m

We are done!

### 4 Answer

Given:

AD = $601 $

CE = $261 $

Let AE = EB = x

BD = DC = y

Using triangle ABD:

AB^{2} + BD^{2} = AD^{2}

$(2x)_{2}+y_{2}=(601 )_{2}$

$4x_{2}+y_{2}=601$ (Equation 1)

Using triangle EBC :

EB^{2} + BC^{2} = EC^{2}

$x_{2}+(2y)_{2}=(261 )_{2}$

$x_{2}+4y_{2}=(261 )_{2}$

$x_{2}+4y_{2}=244$ (Equation 2)

Adding Equation 1 and 2:

4x^{2} + y^{2} + x^{2} + 4y^{2} = 601 + 244

5x^{2} + 5y^{2} = 845

$x_{2}+y_{2}=169$ (Equation 3)

Let us solve for the hypotenuse using the triangle ABC:

AB^{2} + BC^{2} = AC^{2}

(2x)^{2} + (2y)^{2} = AC^{2}

4x^{2} + 4y^{2} = AC^{2}

4(x^{2} + y^{2}) = AC^{2}

Substituting the value of $x_{2}+y_{2}$ from Equation 3:

4(169) = AC^{2}

$AC=4(169) $

AC = 2 × 13 = 26