TriangleABCis a right triangle with heighthto the hypotenuse. The height divides the hypotenuse into segmentsa1 andb1. Prove thath^2 =a1 *b1 without using triangle similarity.

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- Nov 1st 2010, 04:21 PMMATNTRNGRight Triangle Proof
Triangle

*ABC*is a right triangle with height*h*to the hypotenuse. The height divides the hypotenuse into segments*a*1 and*b*1. Prove that*h*^2 =*a*1 **b*1 without using triangle similarity.

- Nov 1st 2010, 04:34 PMArchie Meade
Using Pythagoras' theorem,

$\displaystyle \left(a_1\right)^2+h^2=a^2$

$\displaystyle a^2+b^2=\left(a_1+b_1\right)^2$

$\displaystyle \left(b_1\right)^2+h^2=b^2$

Therefore

$\displaystyle \left(a_1\right)^2+h^2+\left(b_1\right)^2+h^2=\lef t(a_1+b_1\right)^2$

from which the result follows - Nov 1st 2010, 08:25 PMMATNTRNG
If it is not known that triangle

*ABC*is a right triangle but it is known that*h*^2 =*a*1 **b*1, can it be proved that angle*C*must be a right angle? I know that the answer is yes. Any ideas as to how I would go about proving this? - Nov 2nd 2010, 04:03 AMArchie Meade
Yes,

as the non-right-angles are acute, we have

$\displaystyle a_1b_1=h^2\Rightarrow\frac{a_1b_1}{h}=h\Rightarrow \frac{b_1}{h}=\frac{h}{a_1} $

These ratios are the tangents of two angles, which being equal causes the angles to be equal.

Both inner triangles have a right-angle, therefore the third angles are also equal

and the two acute angles sum to 90 degrees. - Nov 2nd 2010, 06:35 PMbjhopperright triangle proof
Hi guys,

Solution 0f followup question using Pythagorean Theorem and same side ids given h^2 =a1 x b1

h^2 + a1^2 =a^2

h^2 + b1^2 =b^2

2h^2 + a1^2 + b1^2 =a^2 + b^2

2a1xb1 + a1^2 + b1^2 = a^2 +b^2

(a1 + b1 )^2 = a^2 + b^2

triangle is a right triangle