Proof (1 Viewer)

[=)(=]

△ABC is isosceles. (AB = BC).

AB is produced to D, and BE is drawn through B parallel to the base AC.

Prove that BE bisects ∠CBD.

Pro

 

[Drongoski]

/_ BAC = /_ DBE = @ say (corresp angles, AC//BE)

/_ EBC = /_ BCA = # say (alt. angles, AC//BE)

Now /_ BAC = /_ BCA (base angles of isosceles triangle ABC)

.: @ = #

i.e. /_ DBE = /_ EBC

.: BE bisects angle CBD.

QED

 

[5uckerberg]

There are many ways to do this Q. Drongoski pointed out one way but I have a different way

alternate angles on lines 1
angles in an isosceles triangle. 2
external angle is the sum of two internal angles. 3
Using 1 and sub it into 3 we will have
As such we will see that BE bisects

 

[CM_Tutor]

It could also be made needlessly complex by asking it in MX1 by vector methods