## Elementary Geometry for College Students (6th Edition)

We consider a rectangle. where the sides are a, b, c, and d and where the diagonals are A and B. Thus, we obtain: $A^2 = a^2 + b^2$ And: $B^2 = c^2 + d^2$ Adding these gives: $A^2 + B^2 = a^2 + b^2 +c^2 + d^2$ Thus, the proof is completed.
In a $parallelogram$ ABCD, we have to prove that diagonals $DB^{2} + AC^{2} = AB^{2}+BC^{2}+DC^{2} + AD^{2}$. 1-Drawing two altitudes AT where T is a point on segment DC , and BV where V is an exterior point on the opposite line. 2-proving triangle DAT = triangle BCV by LH theorem, therefore AT= BV, DT= CV. 3- let AB= DC= b , AD= BC= a, and let TC = c. 4- applying Pythagoras theorem on triangle BVD, $BD^{2} = DV^{2} + BV^{2}$ And applying it on triangle ATC $AC^{2} = AT^{2} + TC^{2}$ 5- writing segments in term of a,b and c Segment DV= AB + DC -TC = 2b-c. ( segment subtraction method ) Segment $AT^{2}=BV^{2}= a^{2} - ( b-c )^2$ 6- by substituting in the Pythagoras theorems above such that $BD^{2} = (a^{2} - ( b-c)^2 ) + ( 2b-c)^2$ And $AC^{2}= (a^2 -(b-c)^2) + c^{2}$ So $BD^2 + AC^2 = 2 ( a^2 -(b-c)^2) + c^2 + ( 2b-c)^2=.....= 2( a^2 + b^2) = a^2+b^2+a^2+b^2$ Since a= AD= BC and b= AB= DC. So we prove that $AB^2+ AC^2= AB^2+BC^2+DC^2+AD^2$ $\square$