In a right triangle ABC with AH as height prove that BC, BA, BH form geometric progression.

Question

Let's have a right triangle ABC(A=90 grades) and AH as height. Prove that BC, BA, BH form a geometric progression. Well is saw that they were similar triangles and showed that $$BC/AC=AC/HC=AB/AH$$ How should I continue?

Answer 1

First note that $[\Delta ABC]=(BA)(AC)/2=(AH)(BC)/2 \to BA=(BC){(AH) \over (AC)}$.

Letting $r={(AH) \over (AC)}$ we can say $BA=(BC)r$.

Now we must show $BH=(BC)r^2$.

Note like you did that $\Delta ABC \sim \Delta BAH$ so ${[\Delta BAH] \over [\Delta ABC]}=k^2$ where $k$ is the ratio of their side lengths.$[\Delta BAH]=(BH)(AH)/2$ and $[\Delta ABC]=(BC)(AH)/2$ and $k={(AH) \over (AC)}$ so ${[\Delta BAH] \over [\Delta ABC]}={(BH)(AH)/2 \over (BC)(AH)/2}={(AH)^2 \over (AC)^2} \to {(BH) \over (BC)}={(AH)^2 \over (AC)^2} \to BH=(BC){(AH)^2 \over (AC)^2}=(BC)r^2$. Hence $BC, BA, BH \to BC, (BC)r, (BC)r^2$