---
book: 4
number: 10
id: "IV.10"
kind: "construction"
uses: ["[[book-2/proposition-11]]", "[[book-4/proposition-1]]", "[[book-4/proposition-5]]", "[[book-3/proposition-37]]", "[[book-3/proposition-32]]", "[[book-1/proposition-32]]", "[[book-1/proposition-5]]", "[[book-1/proposition-6]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.10"
license: "CC-BY-SA-4.0"
---

# IV.10

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

## Proof

Let any straight line *AB* be set out, and let it be cut at the point *C* so that the rectangle contained by *AB*, *BC* is equal to the square on *CA*; [[book-2/proposition-11|II. 11]] with centre *A* and distance *AB* let the circle *BDE* be described, and let there be fitted in the circle *BDE* the straight line *BD* equal to the straight line *AC* which is not greater than the diameter of the circle *BDE*. [[book-4/proposition-1|IV. 1]]

Let *AD*, *DC* be joined, and let the circle *ACD* be circumscribed about the triangle *ACD*. [[book-4/proposition-5|IV. 5]]

Then, since the rectangle *AB*, *BC* is equal to the square on *AC*, and *AC* is equal to *BD*, therefore the rectangle *AB*, *BC* is equal to the square on *BD*.

And, since a point *B* has been taken outside the circle *ACD*, and from *B* the two straight lines *BA*, *BD* have fallen on the circle *ACD*, and one of them cuts it, while the other falls on it, and the rectangle *AB*, *BC* is equal to the square on *BD*, therefore *BD* touches the circle *ACD*. [[book-3/proposition-37|III. 37]]

Since, then, *BD* touches it, and *DC* is drawn across from the point of contact at *D*, therefore the angle *BDC* is equal to the angle *DAC* in the alternate segment of the circle. [[book-3/proposition-32|III. 32]]

Since, then, the angle *BDC* is equal to the angle *DAC*, let the angle *CDA* be added to each; therefore the whole angle *BDA* is equal to the two angles *CDA*, *DAC*.

But the exterior angle *BCD* is equal to the angles *CDA*, *DAC*; [[book-1/proposition-32|I. 32]] therefore the angle *BDA* is also equal to the angle *BCD*.

But the angle *BDA* is equal to the angle *CBD*, since the side *AD* is also equal to *AB*; [[book-1/proposition-5|I. 5]] so that the angle *DBA* is also equal to the angle *BCD*.

Therefore the three angles *BDA*, *DBA*, *BCD* are equal to one another.

And, since the angle *DBC* is equal to the angle *BCD*, the side *BD* is also equal to the side *DC*. [[book-1/proposition-6|I. 6]]

But *BD* is by hypothesis equal to *CA*; therefore *CA* is also equal to *CD*, so that the angle *CDA* is also equal to the angle *DAC*; [[book-1/proposition-5|I. 5]] therefore the angles *CDA*, *DAC* are double of the angle *DAC*.

But the angle *BCD* is equal to the angles *CDA*, *DAC*; therefore the angle *BCD* is also double of the angle *CAD*.

But the angle *BCD* is equal to each of the angles *BDA*, *DBA*; therefore each of the angles *BDA*, *DBA* is also double of the angle *DAB*.

Therefore the isosceles triangle *ABD* has been constructed having each of the angles at the base *DB* double of the remaining one. Q. E. F.