# 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.

In the words of Euclid:

To construct an isosceles triangle having each of the angles at the base double of the remaining one.

## 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 $CD$.

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 Triangle has 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 Triangle has 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 proof is Proposition $10$ of Book $\text{IV}$ 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 $ABD$.