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In triangle $ABC,$ $\angle B = 90^\circ.$ Point $X$ is on $\overline{AC}$ such that $\angle BXA = 90^\circ,$ $AX = 5,$ and $CX = 5$. What is $BX$?

 Apr 8, 2024

Best Answer 

 #1
avatar+14943 
+1

What is BX?

 

Angle B is 90°. Angle BXA is 90°.

So BX is the height of the Thales circle with radius 5.

\(\color{blue}\overline {BX}=5\)

 

laugh !

 Apr 9, 2024
 #1
avatar+14943 
+1
Best Answer

What is BX?

 

Angle B is 90°. Angle BXA is 90°.

So BX is the height of the Thales circle with radius 5.

\(\color{blue}\overline {BX}=5\)

 

laugh !

asinus Apr 9, 2024
 #2
avatar+128794 
+1

A

      5

            X

               5

B                 C

 

 

AX =CX  . ..so ....BX  is also an angle  bisector .......so  AB = BC

 

Triangle AXB ≈  Triangle ABC

AX / AB  = AB / AC

5 / AB = AB / 10

AB^2  = 50

AB = 5sqrt (2) = BC

 

BX / AB  = BC / AC

BX / AB = AB  / AC

BX / AB = AB / 10

10BX = AB^2

10 BX = 50

BX = 50 / 10    =  5

 

cool cool cool

 Apr 9, 2024
edited by CPhill  Apr 9, 2024

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