Maths Problem
Posted:
Tue Jun 17, 2008 2:35 pmby siddhartha_dutta
how to solve this:
http://techrecipe.blogspot.com/2008/06/test.htmlHere DL&BM are angle bisectors
and we have to prove that <BMC+<ALD= 1/2 (<ADC+<ABC>)
Re: Maths Problem
Posted:
Sun Aug 31, 2008 7:57 pmby shubham
some easy ways to solve problems
Re: Maths Problem
Posted:
Fri May 15, 2009 8:49 amby pspoornima
Hello Siddharth,
Please find the answer for your question.
In Parallelogram ABCD,
AB II DC
AL II MC (Extensions of AB and DC Rescpectively)
Ang ABM = Ang BMC (Alternate Angles) (1)
Ang ALD= Ang LDC ( Alternate Angles) (2)
Ang ABM = 1/2 Ang ABC ( Given BM is angle bisector of ang ABC)
Ang LDC = 1/2 Ang ADC ( Given DL is angle bisector of ang ADC)
Ang ABM+LDC = 1/2(Ang ABC+Ang ADC) (3)
But Ang ABM = Ang BMC & Ang LDC =ALD ( from 1 & 2)
Therefore
Ang BMC+ Ang ALD= 1/2(Ang ABC+ Ang ADC) ( from 1, 2 & 3)
Thanks & Regards,
Poornima.