MHB Prove: (AC)² = (AD)² + (AB)·(CD)

Thread starter Albert1
Start date Oct 23, 2013
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Ac

Oct 23, 2013

Albert1

A trapezoid ABCD ,AB//CD ,AD=BC

Prove :

$AC^2=AD^2+(AB).(CD)$

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Oct 30, 2013

Albert1

Re: Prove :AC^2=AD^2+(AB).(CD)

Albert said:

A trapezoid ABCD ,AB//CD ,AD=BC

Prove :

$AC^2=AD^2+(AB).(CD)$

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Thread 'Midpoint(s) of the unbounded number line'

Just chatting with my son about Maths and he casually mentioned that 0 would be the midpoint of the number line from -inf to +inf. I wondered whether it wouldn’t be more accurate to say there is no single midpoint. Couldn’t you make an argument that any real number is exactly halfway between -inf and +inf?

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Thread 'Arithmetic and our place value system'

A power has two parts. Base and Exponent. A number 423 in base 10 can be written in other bases as well: 1. 4* 10^2 + 2*10^1 + 3*10^0 = 423 2. 1*7^3 + 1*7^2 + 4*7^1 + 3*7^0 = 1143 3. 7*60^1 + 3*60^0 = 73 All three expressions are equal in quantity. But I have written the multiplier of powers to form numbers in different bases. Is this what place value system is in essence ?

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MHB Prove: (AC)² = (AD)² + (AB)·(CD)

Attachments

Thread 'Trigonometry problem of interest'

Thread 'Midpoint(s) of the unbounded number line'

Thread 'Arithmetic and our place value system'

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