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

Question

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 $ET_2$