## Friday, April 10, 2015

### Geometry Problem 1109: Right Triangle, External Squares, Catheti, Cathetus, 45 Degree, Angle

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

1. From Problem 1108,
BM=BN

Thus
∠P = ∠A - ∠PNA = ∠A - 45°

2. BM / a =c / (c + a), BN / c = a / (a + c)
so BM = BN, ∠PNA = ∠BNM = 45° etc.

3. We note: AB = c, AC = b, BM = x, BN = y, angle PCB = C, angle CAB = A.

Because ABDE is square, we have that AB = BD = DE = c.
Because BCFG is square, we have that BC = FG = GB.

If triangle ABM is similar with triangle AGF (BM is parallel with GF), then AB/AG = BM/GF or in other words c/(c+a) = x/a. From last equality, we obtain that x = ac/(a+c) [1]

If triangle CBN is similar with triangle CDE (BN is parallel with DE), then CB/CD = BN/DE or in other words a/(a+c) = y/c. From last equality, we obtain that y = ac/(a+c) [2]

From [1] and [2], it resulte that x = y. So triangle BMN is right isosceles because the measure of angle MBN is 90 degree. So, the measure of angle BMN is 45 degree.

Because the angle BMN is exterior for the MCP triangle, then the sum of the measures of the angles P and PCM is equal to the measure of the angle BMN, in other words P + C = 45 degree [3].

But A + C = 90 degree and substituting the expression of C in [3], we obtain P + 90 degree - A = 45 degree, from where P = A + 45 degrees - 90 degrees or P = A - 45 degrees. QED