What is confounding me is that it’s in the book! I feel like I’m missing something. The diagram itself seems impossible to draw based on the assumptions made. If the purple triangle is a right-triangle, and AB = CH then AC and EC would be the same line wouldn’t they?! (As would BC and DC). I think I’m losing my mind.

Look at the big triangle, area G, and the two smaller triangles, E and F. Their sides are proportional so their areas are “proportional squared”.
This means that as the areas of the two little ones have total area equal to the area of the big triangle we get square on hypotenuse equals sum of squares on little sides. (proportion k but it doesn’t matter).

Actually the proof is correct, but missing some essential details.
1. Don’t assume AB bisects HC: it only does this when, as you appear to assume, it goes through the midpoint of AB, the triangle is right-isosceles and then the lines indeed coincide. In all other cases, the purple triangle ABH is smaller in area than triangle ABC. (I also made this error at first because the diagram visually suggests a congruence that isn’t there.)
Instead, construct the square, which I’ll label PQRS, starting with a parallel to AB through H which intersects a perpendicular to AB through B at point P, and the perpendicular to AB through A at Q. Continue to construct the square marking R on QA produced and S on PB produced. AB cuts the square into two pieces, but (usually) they aren’t the same size.
It should now be clear why the area of the triangles sum to half h^2 as the purple triangle is still half the (smaller) top rectangle and triangle ABC half the (larger) lower one.
2. Now drop a parallel to BC through H and label its intersection with HC as point T. Note this makes a parallelogram CBHT which is equal in area to the square a^2 since they share a base in BH and have equal heights. Halving this parallelogram gives us area of triangle CBH = 0.5a^2.
3. Similarly construct a parallel to AC through H and allow it to meet CD produced at V. The second parallelogram CVHA is equal in area to b^2 as above and thus area of triangle CHA = 0.5b^2.
The rest of the proof follows easily, I think.
I’ve constructed it in Geogebra and have a screenshot of the improved diagram which this margin is too small to contain(!) ;o)

This is a fake.

What is confounding me is that it’s in the book! I feel like I’m missing something. The diagram itself seems impossible to draw based on the assumptions made. If the purple triangle is a right-triangle, and AB = CH then AC and EC would be the same line wouldn’t they?! (As would BC and DC). I think I’m losing my mind.

Book is wrong !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Look at the big triangle, area G, and the two smaller triangles, E and F. Their sides are proportional so their areas are “proportional squared”.

This means that as the areas of the two little ones have total area equal to the area of the big triangle we get square on hypotenuse equals sum of squares on little sides. (proportion k but it doesn’t matter).

Actually the proof is correct, but missing some essential details.

1. Don’t assume AB bisects HC: it only does this when, as you appear to assume, it goes through the midpoint of AB, the triangle is right-isosceles and then the lines indeed coincide. In all other cases, the purple triangle ABH is smaller in area than triangle ABC. (I also made this error at first because the diagram visually suggests a congruence that isn’t there.)

Instead, construct the square, which I’ll label PQRS, starting with a parallel to AB through H which intersects a perpendicular to AB through B at point P, and the perpendicular to AB through A at Q. Continue to construct the square marking R on QA produced and S on PB produced. AB cuts the square into two pieces, but (usually) they aren’t the same size.

It should now be clear why the area of the triangles sum to half h^2 as the purple triangle is still half the (smaller) top rectangle and triangle ABC half the (larger) lower one.

2. Now drop a parallel to BC through H and label its intersection with HC as point T. Note this makes a parallelogram CBHT which is equal in area to the square a^2 since they share a base in BH and have equal heights. Halving this parallelogram gives us area of triangle CBH = 0.5a^2.

3. Similarly construct a parallel to AC through H and allow it to meet CD produced at V. The second parallelogram CVHA is equal in area to b^2 as above and thus area of triangle CHA = 0.5b^2.

The rest of the proof follows easily, I think.

I’ve constructed it in Geogebra and have a screenshot of the improved diagram which this margin is too small to contain(!) ;o)