Let ABC be an acute-angled triangle AD be the bisector of ∠ BAC with D on BC and BE be the altitude from B on AC. Show that ∠ CED > 45º.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Let AB = c ′ and AC = b ′
p.v. of D is 
Also 
. 
= 0
Now cos ∠ CED = 
⇒ cos θ = 
= 
⇒ ( b ′ ) 2 
cos 2 θ = ( c ′ ) 2 
sin 2 θ ⇒ 
= 
× 
⇒ cot 2 θ = 
... (i)
Since A,B,C are all acute.
Hence to prove that 
< 1
or cos C < sin B Now since B + C > 
⇒ B > 
– C
or sin B > cos C ⇒ from equation (i) cot θ < 1 ⇒ θ > 45°
Hence Proved.
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems