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boffeemadrid

Answer:

HC=[tex]\sqrt{3}[/tex]

Step-by-step explanation:

Let HC=x, it is given that AH=3HC, then AH=3x.

Since, from the given figure, ΔABC is similar to ΔBHC and ΔABC is similar to ΔABH.

Therefore, ΔABH is similar to ΔBHC, hence using the similarity conditions,

[tex]\frac{HC}{BH}=\frac{BH}{AH}[/tex]

[tex]\frac{x}{3}=\frac{3}{3x}[/tex]

[tex]3x^{2}=9[/tex]

[tex]x^{2}=3[/tex]

[tex]x=\sqrt{3}[/tex]

Hence, HC=[tex]\sqrt{3}[/tex].


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abidemiokin

The measure of the side HC is √3

Using the similarity theorem of a triangle:

HC/BH = BH/AH

Given the following parameters

BH=3

AH=3HC

Substitute into the expression above;

HC/3 = 3/3HC

3^2 = 3HC^2

9 = 3 HC^2

HC^2 = 3

HC = √3

Hence the measure of the side HC is √3

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