Question
Question: If D is the mid-point of side BC of a triangle ABC and AD is perpendicular to AC, then...
If D is the mid-point of side BC of a triangle ABC and AD is perpendicular to AC, then
A
3b2 = a2 – c2
B
3a2 = b2 – 3c2
C
b2 = a2 – c2
D
a2+ b2= 5c2
Answer
3b2 = a2 – c2
Explanation
Solution
From the right-angled triangle CAD, we have
cos C =
Ž a2b=2aba2+b2−c2
Ž a2 + b2 – c2 = 4b2
Ž a2 – c2 = 3b2