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In right triangle \(ABC\)\(AC = BC\) and \(\angle C = 90^\circ\). Let \(P\) and \(Q\) be points on hypotenuse \(\overline{AB}\), as shown below, such that \(\angle PCQ = 45^\circ\). Show that \(AP^2 + BQ^2 = PQ^2.\)

 

Previously posted here.

 Jan 26, 2021

Best Answer 

 #1
avatar+1034 
+3

AC = BC = √50    |    AB = 10    |   CS = 5   |    AP = BQ

 

∠BCQ ≅ ∠ACP           ∠PCQ = 45º           ∠QCS = 22.5º

 

PQ = 2(tan∠QCS * CS)       PQ = 4.142135624

 

AP = BQ = 1/2(AB - PQ)      AP = BQ =  2.928932188

 

PQ2 = 4.1421356242 = 17.15728753

 

AP2 + BQ2 = 2.9289321882 + 2.9289321882 = 17.15728753

 

 Jan 26, 2021
 #1
avatar+1034 
+3
Best Answer

AC = BC = √50    |    AB = 10    |   CS = 5   |    AP = BQ

 

∠BCQ ≅ ∠ACP           ∠PCQ = 45º           ∠QCS = 22.5º

 

PQ = 2(tan∠QCS * CS)       PQ = 4.142135624

 

AP = BQ = 1/2(AB - PQ)      AP = BQ =  2.928932188

 

PQ2 = 4.1421356242 = 17.15728753

 

AP2 + BQ2 = 2.9289321882 + 2.9289321882 = 17.15728753

 

jugoslav Jan 26, 2021
 #2
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0

Hi! Your answer is very clear, but I'm not entirely sure this works since it looks like your solution relies on ∠BCQ ≅ ∠ACP, which may not always be the case.

Guest Jan 27, 2021

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