Solutions to Homework 3 for MATH 128A.

Proofs in circle geometry.
(a)
Solution:

GeoGebra Link Draw the diagonals \(AC\) and \(BD\).

By Proposition III.21, angles standing on the same chord are equal. In particular,

\[ \angle DAC = \angle DBC \qquad \text {and}\qquad \angle CAB = \angle CDB. \]

Thus,

\[ \angle DAB = \angle DAC + \angle CAB = \angle DBC + \angle CDB. \]

In triangle \(BCD\), Proposition I.32 gives

\[ \angle DBC + \angle BCD + \angle CDB = 2\text { right angles}. \]
Therefore,
\[ \angle DAB + \angle BCD = 2\text { right angles}. \]
(b)
Solution:

GeoGebra Link Let \(D\) be the intersection of the angle bisectors of \(\angle A\) and \(\angle B\).

Construct perpendiculars:

\[ DE \perp AB,\quad DF \perp BC,\quad DG \perp AC. \]

Since \(AD\) bisects \(\angle A\),

\[ \angle EAD = \angle DAG, \]
and both \(\angle AED\) and \(\angle AGD\) are right angles. Hence, by AAS,
\[ \triangle AED \cong \triangle AGD, \]
so \(DE = DG\).

Similarly, since \(BD\) bisects \(\angle B\),

\[ \triangle BED \cong \triangle BFD, \]
so \(DE = DF\).

Thus \(DF = DG\).

Now consider right triangles \(DCF\) and \(DCG\). We have

\[ DF = DG, \quad DC = DC. \]
By RASS,
\[ \triangle DCF \cong \triangle DCG, \]
so
\[ \angle FCD = \angle DCG. \]

Hence \(DC\) bisects \(\angle C\).

(c)
Solution:

GeoGebra Link Let two circles be tangent externally at \(A\). Let \(B,C\) be points on the first circle, and let the lines \((AB)\) and \((AC)\) meet the second circle again at \(D\) and \(E\), respectively.

Let \(O_1\) and \(O_2\) be the centers of the two circles. Since the circles are tangent at \(A\), the points \(O_1,A,O_2\) are collinear. Hence the line through \(A\) perpendicular to \(O_1A\) (equivalently, to \(O_2A\)) is tangent to both circles at \(A\).

Let \(t\) denote this common tangent at \(A\).

By Proposition III.32 (the angle between a tangent and a chord equals the angle in the opposite arc), applied to the first circle, we obtain

\[ \angle (t,AB) = \angle ACB. \]

Applying Proposition III.32 to the second circle, we obtain

\[ \angle (t,AD) = \angle AED. \]

Since \(AB\) and \(AD\) lie on the same straight line, the angles \(\angle (t,AB)\) and \(\angle (t,AD)\) are equal. Hence

\[ \angle ACB = \angle AED. \]

Since \(AC\) is a transversal intersecting the lines \((BC)\) and \((DE)\), this implies

\[ BC \parallel DE. \]
GeoGebra Link Construction.
Solution:

To construct the circumscribed circle of triangle \(ABC\):

(a)
Draw a circle centered at \(A\) with radius \(AB\), and a circle centered at \(B\) with radius \(BA\). Let \(E\) and \(F\) be their intersection points. Draw the line \(EF\) (this is the perpendicular bisector of \(AB\)).
(b)
Draw a circle centered at \(A\) with radius \(AC\), and a circle centered at \(C\) with radius \(CA\). Let \(G\) and \(H\) be their intersection points. Draw the line \(GH\) (this is the perpendicular bisector of \(AC\)).
(c)
Let \(D\) be the intersection point of the lines \(EF\) and \(GH\).
(d)
Draw the circle centered at \(D\) with radius \(DA\).

Number of moves:

  • 4 circles (two for each perpendicular bisector),
  • 2 lines (the two perpendicular bisectors),
  • 1 circle (center \(D\), radius \(DA\)).

Total: \(7\) moves.

Reading: the construction of the pentagon.
(a)
Solution:

GeoGebra Link Following Hartshorne Problem 4.1, we construct an isosceles triangle \(ABC\) such that

\[ AB = AC \qquad \text {and}\qquad \angle B = \angle C = 2\angle A. \]
(b)
Solution:

Proposition II.11 (summary). GeoGebra Link

A segment \(AB\) is divided at a point \(H\) such that

\[ AB \cdot BH = AH^2. \]

Proposition III.36 (summary). GeoGebra Link

From a point outside a circle, a tangent and a secant are drawn. Then

\[ (\text {tangent})^2 = (\text {whole secant})(\text {external part}). \]
(c)
Solution:

GeoGebra Link Following Hartshorne Problem 4.2, we construct five equally spaced points on a circle and join consecutive points to obtain a regular pentagon.