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I encountered this problem which I'm not really sure how to solve.

$$AB=1\ \text{unit},\quad CD=\sqrt2\ \text{units},\quad AD=\sqrt2\ \text{units}\quad BD=2\ \text{units},\quad\angle ACB=90^\circ$$

The only other thing I was able to find was the height of triangle $ABD$, which is $\dfrac{\sqrt{7}}4$ units.

I'm supposed to find the length of $AC$ (without using a calculator). Does anyone have an idea how?

1 Answers 1

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Let M be the mid point of AB. Then MCDA is a kite with MD being the symmetry diagonal (ask if you don't see why).You can calculate the length of MD as the median in the triangle ABD according to a well known formula. I get $\frac{\sqrt{11}}{2}$ as the result, but please recheck that, I may have made an error.

Now you know that AC is double the height from A to side MD in triangle AMD. You can calculate that height the same way you did the height from A to side BD in triangle ABD (possibly using Heron's formula), but I assume that the result is "messy" and not some "simple" term.

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    You used Apollonius to find the median right? – 2017-02-01
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    I'm referring to the formula under https://en.wikipedia.org/wiki/Median_(geometry). I didn't know that the formula had a name. – 2017-02-01