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.

SaltyGrandma Jan 19, 2019

#1**+2 **

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

CPhill Jan 19, 2019

#2**+2 **

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

CPhill Jan 19, 2019