# 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

## Prove Euclid's right traingle Theorem 1 and 2: ET_1 =>\overline{BC}^{2} = \overline{AC}*\overline{CH}; $E T {'}_{1} \implies \setminus {\overline{A B}}^{2} = \setminus \overline{A C} \cdot \setminus \overline{A H}$ $E {T}_{2} \implies \setminus {\overline{A H}}^{2} = \setminus \overline{A H} \cdot \setminus \overline{C H}$?

Jan 18, 2017

See the Proof in The Explanation Section.

#### Explanation:

Let us observe that, in $\Delta A B C \mathmr{and} \Delta B H C$, we have,

$\angle B = \angle B H C = {90}^{\circ} , \text{common "/_C=" common } \angle B C H , \mathmr{and} , \therefore ,$

$\angle A = \angle H B C \Rightarrow \Delta A B C \text{ is similar to } \Delta B H C$

Accordingly, their corresponding sides are proportional.

$\therefore \frac{A C}{B C} = \frac{A B}{B H} = \frac{B C}{C H} , i . e . , \frac{A C}{B C} = \frac{B C}{C H}$

$\Rightarrow B {C}^{2} = A C \cdot C H$

This proves $E {T}_{1}$. The Proof of $E T {'}_{1}$ is similar.

To prove $E {T}_{2}$, we show that $\Delta A H B \mathmr{and} \Delta B H C$ are

similar.

In $\Delta A H B , \angle A H B = {90}^{\circ} \therefore \angle A B H + \angle B A H = {90}^{\circ} \ldots \ldots \left(1\right)$.

Also, $\angle A B C = {90}^{\circ} \Rightarrow \angle A B H + \angle H B C = {90}^{\circ} \ldots \ldots \ldots \left(2\right)$.

Comparing $\left(1\right) \mathmr{and} \left(2\right) , \angle B A H = \angle H B C \ldots \ldots \ldots \ldots \ldots . \left(3\right)$.

Thus, in $\Delta A H B \mathmr{and} \Delta B H C ,$ we have,

$\angle A H B = \angle B H C = {90}^{\circ} , \angle B A H = \angle H B C \ldots \ldots \ldots \ldots . \left[\because , \left(3\right)\right]$

$\Rightarrow \Delta A H B \text{ is similar to } \Delta B H C .$

$\Rightarrow \frac{A B}{B C} = \frac{B H}{C H} = \frac{A H}{B H}$

From the ${2}^{n d} \mathmr{and} {3}^{r d} \text{ ratio, } B {H}^{2} = A H \cdot C H$.

This proves $E {T}_{2}$