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Isosceles triangle...

Wilmer

In Memoriam
Mar 19, 2012
376
Code:
                    A

              (51)
                                      
           D
                          
                    E
     
                  (42)        

C        (56)       H       (56)       B
Isosceles triangle ABC, AB = AC, base BC = 112.
D is on AC: line BD crosses height line AH at E.
Results in AD = 51 and EH = 42.
What is the length of height line AH?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Code:
                    A

              (51)
                                      
           D
                          
                    E
     
                  (42)        

C        (56)       H       (56)       B
Isosceles triangle ABC, AB = AC, base BC = 112.
D is on AC: line BD crosses height line AH at E.
Results in AD = 51 and EH = 42.
What is the length of height line AH?
Hi wilmer, :)

Let, \(AE=x\mbox{ and }\angle ADE=\theta\). Using the Pythagorean law for triangle \(ACH\) we get,

\[CD=\sqrt{(x+42)^2+56^2}-51~~~~~~~~(1)\]

Using the law of sines on the triangle \(ADE\) you will get,

\[\sin\theta=\frac{56x}{3570}~~~~~~~~~~~(2)\]

Using the law of sines on the triangle \(BCD\) and using (1) and (2) you will get,

\[x\left(\sqrt{(x+42)^2+56^2}-51\right)=4284+51x\]

Squaring this equation will give you a Quartic equation which has only one positive real solution. I used Maxima to get the answer,

\[x=63\]

This method may not be the most economical way of doing this problem and I would be delighted to see a more elegant method. :)

Kind Regards,
Sudharaka.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Thanks, Mr, Sud; I agree; but you've been of no help (Smile)
I had solved it (also with a darn Quartic!) and was sneakily trying
to see if someone could come up with something "simpler".

Since I hate using SIN or COS, I placed CB on x-axis with C at origin;
let h = AH, so A(56,h).

Letting (x,y) = D's coordinates, I used following equations:
BD's y-intercept is clearly 84; hence BD's equation is: y = (-3/4)x + 84
AC's equation is easier still, with points (0,0) and (56,h) : y = (h/56)x

So I needed to solve:
(56 - x)^2 + (h - y)^2 = 51^2

Getting x and y in terms of h:
x = 4704 / (h + 42)
y = 84h / (h + 42)

And that leads to MY(!) quartic:
h^4 - 84h^3 + 2299h^2 - 481908h + 943740 = 0 (hope it's better than yours)(Thinking)
Which has h = 105 as only "valid" solution (so AE = 63)

And that checks out ok. Makes the equal sides AB and AC = 119 ; also DE = 30.

Surprising to me that this is not easier, with right triangles all over the place!