## Monday, May 19, 2008

### Elearn Geometry Problem 69

See complete Problem 69
Square Inscribed in a Triangle. Level: High School, SAT Prep, College geometry

1. Thalès twice

2. Let the vertices of square are PQRS and BM perpendicular is drawn to BC. area triangle APM=(1/2).s.AM
area of triangle SMC=(1/2).s.MC
area of triangle PMS=(1/2).s.s
area of triangle BPS=(1/2).s.(h-s)
area of whole triangle=(1/2).b.h
now
1/2).s.AM+(1/2).s.MC+(1/2).s.s+(1/2).s.(h-s)=(1/2).b.h
and AM + MC=b
which on simplifying prove the relation

3. With P, Q, R and S vertices of the square and k be the height of the triangle ΔSBR, we have ΔABC ~ ΔSBR. Therefore,

From h/k = b/s, we have:
s = bk/h = b(h-s)/h = (bh-bs)/h
and sh + bs = bh or s = bh/(b+h)

QED

4. Let PQRS be the square.
Area of triangle ABC= Area of triangle BSR + Area of Trapezium ACRS.
(1/2)bh = (1/2)s*(h-s) + (1/2)(b+s)* (h-s)
bh = (h-s)(2s+b)
on simplification we get the result

5. Area of Tr. ABC is comprised of the areas of 3 Tr.s and the square.

So s^2 + 1/2s(h-s) + 1/2s(b-s) = 1/2 bh

From which the result follows

Sumith Peiris
Moratuwa
Sri Lanka

6. Or from similar triangles

(h-s)/h .= s/b and the result follows