# Prove that altitude² = pq?

The following is a question for my math class. I just cannot figure it out.

Given is that: h is the altitude that divides the longest side of this right triangle into p and q.

Question: Prove that h² = pq

I really have no idea what to do. Could anyone push me in the right direction?

EDIT: I'm now this far:

For convenience, lets call the opposite side of the big triangle s, and the adjacent of the big triangle r.

There are three similar triangles:

• Triangle shq
• Triangle hpr
• Triangle sr(p+q)

Now, using the pythagoras theorem to determine the hypotenuse(i.e. p+q) of the big triangle sr(p+q)

$(p+q)^2 = s^2 + r^2$

Now filling in s and r:

$(p+q)^2 = h^2 + q^2 + h^2 + p^2$

$(p+q)^2 = h^4 + p^2 + q^2$

$p+q = h^2$

but... $p + q$ isn't $p*q$, right? Where is my mistake?

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Hint: Can you see similar triangles in your diagram? –  Mark Bennet Nov 15 '12 at 22:33
ever heard of pythagoras theorem? –  user31280 Nov 15 '12 at 22:37
Look at $q/h$ in the 2nd of the three triangles you mention, and at $h/p$ in the third. –  Gerry Myerson Nov 15 '12 at 22:49
That's some funky algebra you're doing. $h^2+h^2$ isn't $h^4$, and how you get from the 2nd last equation to the last, I think I'd rather not know. –  Gerry Myerson Nov 15 '12 at 23:06
As pointed out by Gerry, $\sqrt{a^2 + b^2} \ne a+b$. So you cannot get $p+q = h^2$ from $(p+q)^2 = h^4+p^2+q^2$. Instead use $(p+q)^2 = p^2 + 2pq + q^2$ –  mythealias Nov 16 '12 at 0:03

Note that $\triangle ADC \sim \triangle BDA$. (Why?) This is so since $$\angle{ADC} = \angle{BDA} = 90^{\circ}$$ Further, $$\underbrace{\angle{CAD} = 90^{\circ} - \angle{ACD}}_{\because\text{ }\triangle ADC \text{ is right-angled at }D} = \underbrace{90^{\circ} - \angle{ACB} = \angle{ABC}}_{\because \text{ }\triangle ABC \text{ is right-angled at } A} = \angle{ABD}$$ Hence, we have that $\triangle ADC \sim \triangle BDA$. Hence, the ratio of the corresponding sides must be equal i.e. $$\dfrac{\text{Side opposite to }\angle{ACD}}{\text{Side opposite to }\angle{DAC}} = \dfrac{\text{Side opposite to }\angle{BAD}}{\text{Side opposite to }\angle{DBA}}$$ From this conclude, what you want to.
Similar triangles. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
Use Pythagoras theorem to get $p^2+h^2+q^2+h^2=(p+q)^2.$