Construction of Isosceles Triangle whose Base Angle is Twice Apex

Theorem

It is possible to construct an isosceles triangle such that each of the angles at the base is twice that at the apex.

Construction

Let $AB$ be a straight line.

Construct $C$ on $AB$ such that $AB \cdot BC = AC^2$.

Construct the circle whose center is at $A$ and whose radius is $AB$.

Fit $BD$ into this circle as a chord such that $BD = AC$.

Join $AD$.

Then $\triangle ABD$ is the required isosceles triangle, as $\angle ABD = \angle BAD = 2 \angle BDA$.

Proof

Join $CD$.

Circumscribe circle $ACD$ about $\triangle ACD$.

As $AC = BD$ we have that $AB \cdot BC = BD^2$.

We have that $B$ is outside the circle $ACD$.

From the converse of the Tangent Secant Theorem it follows that $BD$ is tangent to circle $ACD$.

Then from Angles made by Chord with Tangent‎ $\angle BDC = \angle DAC$.

Add $\angle CDA$ to both:

$\angle CDA + \angle BDC = \angle BDA = \angle CDA + \angle DAC$.

But from Sum of Angles of Triangle Equals Two Right Angles we have that:

$(1) \quad \angle BCD = \angle CDA + \angle DAC$

So $\angle BDA = \angle BCD$.

But since $AD = AB$, from Isosceles Triangles have Two Equal Angles $\angle BDA = \angle CBD$.

So $\angle BDA = \angle BCD = \angle CBD$.

Since $\angle DBC = \angle BCD$, from Triangle with Two Equal Angles is Isosceles we have $BD = DC$.

But by hypothesis $BD = CA$ and so $CA = CD$.

So from Isosceles Triangles have Two Equal Angles $\angle CDA = \angle DAC$.

So $\angle CDA + \angle DAC = 2 \angle DAC$.

But from $(1)$ we have that $\angle BCD = \angle CDA + \angle DAC$.

So $\angle BCD = 2 \angle CAD = 2 \angle BAD$.

But $\angle BCD = \angle BDA = \angle DBA$.

So $\angle ABD = \angle BAD = 2 \angle BDA$.

$\blacksquare$

Historical Note

This is Proposition 10 of Book IV of Euclid's The Elements.

Having established in the proof that $CD$ equals $BD$, the construction can be simplified by constructing the circle whose center is at $C$ and whose radius is $AC$, then identifying $D$ as the point at which circle $ACD$ meets circle $ABD$, instead of invoking the somewhat more cumbersome construction that fits $BD$ into the circle $ABD$.