Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 797.

## Thursday, August 23, 2012

### Problem 797: Parallelogram, Angle Bisector, Isosceles and Similar Triangles

Labels:
angle bisector,
isosceles,
parallelogram,
similarity,
triangulo

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(1)

ReplyDeleteSince angle BAE = angle DAE = angle BEA,

thus BA = BE, and so ABE is isosceles.

(2)

Since AB//DC, so

angle BAE = angle CFE & angle ABE = angle FCE

which means ABE and ECF are similar.

Since BC//AD, so

angle FEC = angle FAD & angle FCE = angle FDA

which means ECF and ADF are similar.

Combining (1) & (2),

they are isosceles and similar.

Angle BAE = angle FAB

ReplyDeleteAngle BAE = angle AFB (parallel lines)

Angle BEA = angle FAB ( parallel lines)

So all these four angles are the same and hence the required triangles are isos and similar

Since AE bisector of BAD then ABE is half of a rhombus,

ReplyDeleteso and two others