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Pythagoras Theorem

In Figure 6, ...

Question

In Figure 6, P and Q are the midpoints of the sides CA and CB respectively of ∆ABC right angled at C. Prove that 4(AQ² + BP²) = 5 AB².

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Solution

In the given figure P and Q are the mid points of AC and BC, the we have to prove that

The following figure is given.

Using Pythagoras theorem in, we get

AB² = AC² + BC²…… (1)

Similarly, by using Pythagoras theorem in, we get

AQ² = CQ² + AC²…… (2)

BP² = CP² + BC²…… (3)

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

…… (4)

Now multiply equation (4) by 4, we get

Hence proved.

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Similar questions

4 A Q^{2} = 4 A C^{2} + B C^{2}

4 B P^{2} = 4 B C^{2} + A C^{2}

4 (A Q^{2} + B P^{2}) = 5 A B^{2}

P

Q

C A

C B

A B C

C

4 (A Q^{2} + B P^{2}) = 5 A B^{2}

P

Q

C A

C B

△ A B C

4 {A Q}^{2} = 4 {A C}^{2} + {B C}^{2}

4 {B P}^{2} = 4 {B C}^{2} + {A C}^{2}

4 ({A Q}^{2} + {B P}^{2}) = 5 {A B}^{2}

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$

P

Q

C A

C B

Δ A B C

C

(A Q^{2} + B P^{2}) = (A B^{2} + P Q^{2}) .

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