Edited by Richard Walker, Thursday, 25 July 2024, 00:41
Consider Fig. 1, which shows squares with sides a and b resting on a bigger square of side a + b. R, Q and S are the centres of the respective squares. Then line segments RS and PQ have equal length and are at right angles.
Proof: In Fig. 2 the base and vertical side of the right-angled triangle shown have length a + b/2 and b/2 respectively.
In Fig.3 for the second right-angled triangle we can work out the length of the base as (a + b)/2 - a/2 = a/2 + b/2 - a/2 = b/2 and the length of the vertical side as a/2 + (a + b)/2 = a/2 + a/2 + b/2 = a + b/2.
So both right-angled triangles have sides of a + b/2 and b/2 and an included angle of 90°, which means they are congruent, i.e. identical. It follows immediately that the hypotenuses PQ and RS are equal. Moreover the sides of length a + b/2 in each triangle are at right angles, and so the same is true of each pair of corresponding sides in the two triangles and PQ must therefore be perpendicular to RS.
A Nice Geometry Problem
Consider Fig. 1, which shows squares with sides a and b resting on a bigger square of side a + b. R, Q and S are the centres of the respective squares. Then line segments RS and PQ have equal length and are at right angles.
Proof: In Fig. 2 the base and vertical side of the right-angled triangle shown have length a + b/2 and b/2 respectively.
In Fig.3 for the second right-angled triangle we can work out the length of the base as (a + b)/2 - a/2 = a/2 + b/2 - a/2 = b/2 and the length of the vertical side as a/2 + (a + b)/2 = a/2 + a/2 + b/2 = a + b/2.
So both right-angled triangles have sides of a + b/2 and b/2 and an included angle of 90°, which means they are congruent, i.e. identical. It follows immediately that the hypotenuses PQ and RS are equal. Moreover the sides of length a + b/2 in each triangle are at right angles, and so the same is true of each pair of corresponding sides in the two triangles and PQ must therefore be perpendicular to RS.