Prove Euclid's right traingle Theorem 1 and 2: #ET_1 =>\overline{BC}^{2} = \overline{AC}\overline{CH};# #ET'_1 =>bar(AB)^{2} =bar(AC)bar(AH)#; #ET_2 =>barAH^{2} = \overline{AH}*\overline{CH}#? ![enter image source here](https

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Prove Euclid's right traingle Theorem 1 and 2: #ET_1 =>\overline{BC}^{2} = \overline{AC}\overline{CH};#
#ET'_1 =>\overline{AB}^{2} = \overline{AC}\overline{AH}#
#ET_2 =>\overline{AH}^{2} = \overline{AH}*\overline{CH}#?

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Answer 1 · 2017-01-18T17:18:47

See the Proof in The Explanation Section.

Let us observe that, in #Delta ABC and Delta BHC#, we have,

# /_B=/_BHC=90^@, "common "/_C=" common "/_BCH, and, :.,#

# /_A=/_HBC rArr Delta ABC" is similar to "Delta BHC#

Accordingly, their corresponding sides are proportional.

#:. (AC)/(BC)=(AB)/(BH)=(BC)/(CH), i.e., (AC)/(BC)=(BC)/(CH)#

#rArr BC^2=AC*CH#

This proves #ET_1#. The Proof of #ET'_1# is similar.

To prove #ET_2#, we show that #Delta AHB and Delta BHC# are

similar.

In #Delta AHB, /_AHB=90^@ :. /_ABH+/_BAH=90^@......(1)#.

Also, #/_ABC=90^@ rArr /_ABH+/_HBC=90^@.........(2)#.

Comparing #(1) and (2), /_BAH=/_HBC................(3)#.

Thus, in #Delta AHB and Delta BHC,# we have,

#/_AHB=/_BHC=90^@, /_BAH=/_HBC.............[because, (3)]#

#rArr Delta AHB" is similar to "Delta BHC.#

#rArr (AB)/(BC)=(BH)/(CH)=(AH)/(BH)#

From the #2^(nd) and 3^(rd)" ratio, "BH^2=AH*CH#.

This proves #ET_2#