# Problem Solving #3

“A square is constructed as shown. Point A is the midpoint of a side. Prove that triangle BCD is isosceles (and not equilateral)”

If you don’t want to be swayed by my thinking, solve it now rather than reading ahead.

I like proof questions because they give you a little comfort blanket – you know the answer before you start. I stared at the diagram a little while, explaining internally how each line is constructed as a kind of mental warm up I suppose. Then I thought to myself “I’ll try and find the side lengths”. I think I was swayed to pursue sides rather than angles because I identified this triangle fairly quickly and figured it’d probably be useful:

I gave it side lengths of 2 and 1 for easy numbers, and so the hypotenuse must be

$\sqrt{5}$

Then I focused for a little while on this red triangle:

I hoped it’d be similar, but couldn’t convince myself that it was by using side lengths. At this point I only know the hypotenuse is 2.

So I thought maybe I could with angles.

I don’t know the values for the red and green angles. I could work them out as I know the side lengths, but I don’t want to yet. Instead, I noted that the green and red angles sum to 90 degrees. Therefore, this angle must also be ‘red’:

And as angles in a triangle sum to 180, and we have a 90 degree angle at point C, then the third angle in the red triangle must be ‘green’, so it’s a similar triangle to the one I started with. Phew!

Why is that useful? Well, now I can calculate side DC using similarity.

$\frac{DC}{2}=\frac{2}{\sqrt{5}}$

$DC = \frac{4\sqrt{5}}{5}$

Well, I’ve proved that triangle BCD isn’t equilateral at least.

Now to find side BC…

I know that angle BDC must also be ‘red’:

I could just use the Cosine rule and find side BC to be 2 or the surd for CD (I assume it would come out at 2 by inspection), but then I’d have to actually work out those angles, and it was late and I needed to get to work early because I’ve been locked out of my work emails all weekend (stupid password change requirements are so inconvenient). I had one last look before bed and thought ‘maybe there’s a way to show that angle DCB is also ‘red’…

The next morning I ate some toast (this is a theme, maths and toast. I highly recommend it). and stared at this diagram:

I stared for a good long while, thinking of that angle next to the right angle. My mind got a bit confused with itself as one of my unwritten approaches in my head was ‘assume it IS red, then what else’, and I found myself in a loop of using something else that would be true if it WAS red, to show that it was red… that doesn’t work in maths sadly.

I even sketched this to see if it would jolt any thoughts out of my brain:

It didn’t. Not that I’m consciously aware of anyway. As I brushed the crumbs from my face and sipped my obligatory morning mug of Yorkshire Tea, an idea finally struck.

I focused again on the red triangle:

It’s a right-angled triangle, so I can circumscribe it. Would that help?

It was a better idea than no idea. A few more details can be deduced now:

So the top line is a tangent, and the midpoint on the side of the square is the centre of the circle…

And the dotted line is a radius, so now I have another pair of red and green angles… i feel like i’m making progress now, but I couldn’t quite make what felt like a final move to unlock the proof. So i drew a second circle as I lost hope that the first would be sufficient on its own (perhaps it is and I missed something):

Notice I’ve ditched that tangent, it’s not important. At last, from this second circle I can happily convince myself that everything works, and the original triangle BCD is isosceles:

I know the two green angles are green from the previous diagram, and i know the dots A, B, C, D & F all lie on the new circle because …

Triangle BDF forms a right triangle for the new circle drawn.

Triangle DBA forms a right triangle with the same height, so the midpoint of its hypotenuse is the same as for BDF – so it’s the same circumscribed circle (so A lies on the circle from BDF)

We know ECD is 90 degrees, so DCA is 90 degrees, so that can be circumscribed.

DA is the hypotenuse of DBA and DCA so all three right triangles have the same circle circumscribing them all. I’m now convinced C lies on both circles.

That’s important because it tells us that BFC is also circumscribed by the same circle, and so FCB is 90 degrees and so DCB is… RED!

I have a feeling I might not have needed both circles for my solution. But I’ve moved on.

Here is the solution approach provided with the problem:

## 4 thoughts on “Problem Solving #3”

1. I didn’t look !
How about a line from B parallel to AC ?

2. Cool solution with the circles. I used a less elegant approach using similar triangles and Pythagoras. But it was fun.
Using the original lettering in the problem let the top left of the square be E. and the top right corner of the square be F.
So AE = sqrt(5) by Pythagoras
Triangles EFA and DCE are similar.
So CD = 4/sqrt(5) by similar triangles
Drop a perpendicular from C to BD meeting BD at P. This is the key step.
Triangle CPD is also similar to EFA and DCE.
CD = 4/sqrt(5) is the hypotenuse of CPD so DP = 4/5 and PC = 8/5 by similar triangles
But BD=2 so BP = 2 – 4/5 = 6/5.
Triangle BPC is similar to a 3, 4, 5 triangle. BP = 6/5; PC = 8/5 so BC = 10/5 = 2 (Pythagoras).
So we have all three sides of triangle BCD namely BD = BC = 2 and CD = 4/sqrt(5) QED

3. This problem also works if you have a rectangle instead of a square – in other words keep the height as 2 units and A is half way up the right hand side but BD can be any length you like.

• Although is the width is sqrt(3) you do get an equilateral triangle.