Similarity Proofs (1 Viewer)

[Lemiixem]

ABCD is a rectangle.
Prove that TriangleBFC is similar to TriangleDCE.

 

[Green Pup]

Angle FBC=CDE=90 D (ABCD is a rectangle)

tanFCB=12/4
FCB=71 D 34 M
Angle BFC=180-C-90 (Angle sum of triangle)
Angle BFC=18 D 26 M

Angle DCE=180-90-C
Angle DCE=18 D 26 M=Angle BFC
Angle CED=180-90-DCE
Angle CED=71 D 34 =Angle FCB

Therefore: TriBFC is similar to TriDCE (equiangluar)

 

[Aesytic]

in triangles BFC, AFE,
angle AFE is common
angle FAE = 90 (angles of a rectangle are right angles)
similarly, angle ABC = 90
.'. angle FBC = 180 - angle ABC (adjacent supplementary angles)
= 90
.'. angle FBC = angle FAE
.'. triangle BFC is similar to triangle AFE (equiangular)
.'. angle FCB = angle CED (corresponding angles in similar triangles are equal)
In triangles BFC, DCE,
angle ADC = 90 (angles in a rectangle are right angles)
.'. angle CDE = 180 - angle ADC (adjacent supplementary angles)
= 90
angle FBC = 90 (proven before)
.'. angle CDE = angle FBC
angle FCB = angle CED (proven before)
.'. triangle BFC is similar to triangle DCE (equiangular)