+83 An angle bisector AC divides a trapezoid ABCD into two similar triangles △ABC and △ACD. Find the perimeter of this trapezoid if the leg AB=9 cm and the leg CD=12 cm.
The perimeter of △CDE is 55 cm. A rhombus DMFN is inscribed in this triangle so that vertices M, F, and N lie on the sides
CD, CE, and DE respectively. Find CD and DE if CF=8 cm and EF=12 cm.
+129933 An angle bisector AC divides a trapezoid ABCD into two similar triangles △ABC and △ACD. Find the perimeter of this trapezoid if the leg AB=9 cm and the leg CD=12 cm.
Let AC bisect A
Then angle CAD = angle CAB
And angle ACD = angle BAC
So...triangle ACD is similar to triangle CAB
So
CD/AB = AD/BC
12/9 = AD/BC
But angle ACD = angle CAB = angle CAD = angle ACB
So AD = CD
And BC = AB
So AD = 12
And BC = 9
So....the perimeter of the trapezoid is AB + BC + CD + AD = 9 + 9 + 12 + 12 = 42
+129933 Here's a pic :
Since MCFN is a rhombus....then FN is parallel to CM
But FN is also parallel to DC
And CE is a transversal cutting parallel segments which makes coresponding angles EFN and ECD equal
And angle FEN = angle CED
So....by AA congruency, triangle NFE is similar to triangle DCE
Then FN / FE = CD /CE
And FN = CF = 8
And FE = 12
And CE = CF + EF = 20
Therefore
8 / 12 = CD / 20
2 / 3 = CD /20
CD = 40/3
Therefore DE = 55 - CE - CD = 55 - 20 - 40/3 = 65/3
