[Math] Circumscribed Circle of an Isosceles Triangle

circleseuclidean-geometrygeometry

Let $o$ be the circumscribed circle for an isosceles triangle $ABC$ with $|AB|=|AC|$. Let $D$ be a point on the side $BC$ and let $E$ be the second intersection point of the line $AD$ with $o$. I need to prove that $|AB|^2 = |AE||AD|$

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Any hints?

Thanks

Best Answer

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The two triangles $ADH$ and $AFE$ are similar, so $$ \frac{AD}{AF}=\frac{AH}{AE}, $$

or, equally: $AD\cdot AE = AH\cdot AF$.

Now, applying the first Euclid's theorem to the triangle $ABF$, you can write:

$$ AH\cdot AF = AB^2, $$

and, by combining the two equations:

$$ AD\cdot AE = AH\cdot AF= AB^2. $$

Better solution:

The two triangles $ABD$ and $ABE$ are similar, so

$$ \frac{AB}{AE}=\frac{AD}{AB}, $$

then $AB^2 = AE\cdot AD$.

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