SOLUTION: In the given fig.AB//CD and P is any point. prove that angle ABP+angle BPD + angle CDP=360^0. (the fig. looks something like this)<pre> B A<----------- .------

Algebra ->  Geometry-proofs -> SOLUTION: In the given fig.AB//CD and P is any point. prove that angle ABP+angle BPD + angle CDP=360^0. (the fig. looks something like this)<pre> B A<----------- .------      Log On


   

Question 1041410: In the given fig.AB//CD and P is any point. prove that angle ABP+angle BPD + angle CDP=360^0.
(the fig. looks something like this)
                B
  A<------------.-------------------->
.................\               
..................\                
...................\                   
....................\P                    
..................../                    
.................../                   
................../                  
................./                 
 C<-------------.------------------->
                D

I HOPE THE FIG. IS UNDERSTOOD
I DONT KNOW HOW TO SEND TH IMAGE ON THIS...I ACTUALLY WANTED TO HOW HOW I DID THIS SUM..PLEASE HELP!
THE DOTS ARE NOTHING THE FIGURE WASNT COMING WITHOUT THEM.
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
It has 5 sides, so the sum of the interior angles = 540 degs
-----
The sum of the angles at A and C = 180 degs since AB is parallel to CD.
That leaves 360 degs for the 3 other angles.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
In the given fig.AB//CD and P is any point. prove that
angle ABP+angle BPD + angle CDP=360^0.
(the fig. looks something like this) 
                B
  A<------------.-------------------->
.................\               
..................\                
...................\                   
....................\P                    
..................../                    
.................../                   
................../                  
................./                 
 C<-------------.------------------->
                D
Draw AC

                B
  A<------------.
   |.............\               
   |..............\                
   |...............\                   
   |................\P                    
   |................/                    
   |.............../                   
   |............../                  
   |............./                 
 C<-------------.
                D

Then figure ACDPB is a pentagon, a 5-sided
polygon.  The formula for the sum of the
interior angles of an n-sided polygon is

(n-2)180°

Since n=5, The sum of angles 
A+C+D+P+B = (5-2)(180°) = 3(180°) = 540°.

The sum of Angles A+C = 180° because they are
angles on the same side of transversal AC cutting
parallel lines AB and CD, and so they are supplementary.

A+C+D+P+B = 540°
A+C       = 180°
---------------=
    D+P+B = 360°   Subtracting equals from equals.

[Notice I just labeled the angles with one letter 
instead of 3, but I erased the extensions of AB and
CD, so you could tell what angles I meant.]

Edwin