1o Trójkąty APB i DPC mają równe podstawy oraz różne wysokości.
h1 + h2 = a
| 1 | ||
PAPB = | * b * h2
| |
| 2 |
| 1 | ||
PDPC = | * b * h1
| |
| 2 |
| 1 | 1 | |||
PAPB + PDPC = | * b * h2 + | * b * h1
| ||
| 2 | 2 |
| 1 | ||
PAPB + PDPC = | * b(h2 + h1)
| |
| 2 |
| 1 | ||
PAPB + PDPC = | * a * b
| |
| 2 |
| 1 | ||
PAPD = | * a * h4
| |
| 2 |
| 1 | ||
PBPC = | * a * h3
| |
| 2 |
| 1 | 1 | |||
PAPD + PBPC = | * a * h4 + | * a * h3
| ||
| 2 | 2 |
| 1 | ||
PAPD + PBPC = | * a(h4 + h3)
| |
| 2 |
| 1 | ||
PAPD + PBPC = | * a * b
| |
| 2 |
należą się dwa jabłka