If A and B are acute angles such that Cos A = Cos B, then show that A = B


Let us consider a triangle ABC in which CD ⊥ AB



 


It is given that


 


Cos A = Cos B


In Δ ADC, cos A = AD/AC


In Δ ADB, cos B = BD/BC


⇒ AD/AC = BD/BC ..(i)


 


We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP, as shown below:



 


Now, From equation (i), we obtain


⇒ AD/AC = BD/BC .


Rewriting we get, (BC = CP)


⇒ AD/AC = BD/CP ..(i)


 


Now, if we join, B and P to get BP, as shown, below:



Then, by using the converse of Basic proportionality theorem we get,


Now, CD||BP


⇒ ∠ACD = ∠CPB (Corresponding angles) ..(iii)


⇒ ∠BCD = ∠CBP (Alternate interior angles) ...(iv)


And, by construction, we have BC = CP


∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) ...(v)


From equations (iii), (iv), and (v), we obtain


∠ACD = ∠BCD ..Eq. (vi)


Now, In ΔCAD and ΔCBD,


∠ACD = ∠BCD [Using equation equation (vi)]


∠CDA = ∠CDB (Both 90°)


Therefore, the remaining angles should be equal


⇒ ∠CAD = ∠CBD


⇒ ∠A = ∠B


 

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