I don't know the Right Triangle Altitude Theorem.
I think that's what I need to solve the problem. The reason why I'm not doing a simple Google search is because other than the actual way to solve this, I want to also know how I can remember it, or in other words, a logical explanation which will help me remember this theorem. I keep forgetting it!
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You don't necessarily need to use or remember the Right Triangle Altitude Theorem to solve this problem. Lets see how we can get to the theorem using simple algebraic operations.
The right triangle altitude theorem states that the altitude drawn from the vertex of right angle to the hypotenuse is equal to the geometric mean of the two segments of hypotenuse.
i.e., `CD = sqrt(AD*BD)`
or, `CD^2 = AD*BD`
Now, lets do this by simple algebraic operations.
Using the Pythagoras Theorem on triangles ADC, BDC and ACB, we get
AB^2 = AC^2 + CB^2
AC^2 = AD^2 + CD^2
CB^2 = CD^2 + BD^2
substituting the values of AC^2 and BC^2 in the first equation, we get
AB^2 = AD^2 + CD^2 + CD^2 + BD^2 = AD^2 + 2 CD^2 + BD^2
or, AB^2 = (AD + BD)^2 = AD^2 + BD^2 + 2 AD*BD = AD^2 + 2CD^2 + BD^2
we end up with, AD*BD = CD^2
which is exactly what the right angle altitude theorem states.
since we are given the values of CD = 6 and AD/AB = 1/5, we can solve for BD.
AB = AD + BD
so, AD/AB = AD/(AD+BD) = 1/5
or, 5 AD = AD + BD
or, AD = BD/4
substituting the value of AD in right triangle altitude theorem, we get
CD^2 = 6^2 = 36 = (BD/4)* BD = BD^2 / 4
or, BD = `sqrt (36*4) = sqrt(144) = 12`
This is how you can remember how to get to right triangle altitude theorem step-by-step, without needing to memorize it.
Hope this helps.
I had explained the problem on a paper .....hope this work will help you.
Please check the attachments for the simple steps answer.
I am sure you can remember this.
Remember when ever ratios comes into picture in Triangles, then using similarity or congruency of triangles will solve the problem easily.
Please learn all the similarity properties, this will help you a lot and makes the solution simple :)
This is simple :
Given `(AD)/(AB) =1/5`
From Delta ABC
`=> (AD)/(AD+DB) =1/5`
`=> (AD+DB)/(AD) =5/1 `
`=> (AD)/(AD) + (DB)/(AD) = 5/1 `
`=> 1+ (DB)/(AD) = 5 `
`=> (DB)/(AD) = 4 `
`=> (DB)/4 = AD `
In `Delta ADC & Delta BCD `
Angle ADC & Angle CDB = `90^0 `
CD=DC IS THE COMMON SIDE
by side - angle similarity we get
`Delta ADC ~~ Delta CDB `
=> by ratio of sides property , we get
`(AD)/(CD)= (DC)/(DB) ` [ this is the same we obtain in Right Triangle Altitude Theorem]
`=> AD*DB = CD^2 `
`=> (DB)/4*DB = 6^2`
`=> DB^2 = 6^2 * 4 = 6^2 * 2^2 `
`=> DB = BD = 6*2 =12 `
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