Triangle ABC is isosceles with AB=BC. There is a point D on AC such that BD = DC and AB=AD.
What is the measure of angle BAC?
(In reply to
answer by K Sengupta)
< BAC = < BCA = p (say)
Since BD=DC it follows that, triangle CDB is also isosceles.
So, < DBC = < BCD = p
Now, suppose < ABD = q (say)
Then < ABC= < ABD + < DBC = q+p
But, < ABC = 180 - 2p
Thus p+q=180-2p
=> 3p+q = 180 .............(i)
Again, AB=AD
So, triangle ABD is isosceles
and, < ABD = < ADB = q
Now, < ADB + < BAD + < ABD =180
=> q+p+q =180
=> p+2q = 180 ........(ii)
Solving for p and q from (i) and (ii), we have:
p =36, and q=72
Consequently, < BAC = 36 degrees.
Explanation to Puzzle Answer
