# Basic Geometry (Similar Triangles)

I am given with a Triangle $$ABC$$ which is inscribed in circle $$\omega$$. The tangent lines to $$\omega$$ at $$B$$ and $$C$$ meet at $$T$$. Point $$S$$ lies on ray $$BC$$ such that $$AS$$ is perpendicular to $$AT$$. Points $$B_1$$ and $$C_1$$ lies on ray $$ST$$ (with $$C_1$$ in between $$B_1$$ and $$S$$) such that $$B_{1}T=BT=C_{1}T$$. I need to prove that the triangles $$ABC$$ and $$AB_1C_1$$ are similar to each other.

• Maybe it means $B_1T=BN=C_1S$? Show please your attempts. – Michael Rozenberg Jan 22 at 7:51
• We have $B_1T=BT=CT=C_1T$. But $B,C$ are the only two points on the line $ST$ which are a distance $BT$ from $T$. So if $B_1,C_1$ are distinct then they must coincide with $B,C$. I think you have a typo somewhere in the question. – almagest Jan 22 at 10:00

Here is a rewritten version of the OP with a picture. The order of "inventing" the points (in the solution using inversion) is the same as the order of the "inverted" points in a lemma. The lemma is simple, extracts the essence and gives the bridge for the solution. (Writing the lemma is not so simple, points have to be constructed in the right order.)

Proposition: There are given two circles with centers $$O$$ and $$T$$, which intersect orthogonally in two points, $$B$$ and $$C$$. The two circles are denoted by $$(O)$$, $$(T)$$, using their centers, if there is no danger of confusion.

Let $$U=OT\cap BC$$ be the mid point of the segment $$BC$$.

Let $$S$$ be a point on the ray = half-line $$BC$$, so that the circle with diameter $$ST$$ intersects $$(O)$$ in two points, $$A$$ and $$a$$. Here $$A$$ is on the bigger arc $$\overset\frown{BC}$$.

Let $$B_1,C_1$$ be the intersections of the line $$ST$$ with the circle $$(T)$$ , so that the points $$B_1,T,C_1,S$$ appear in this order on the line.

Then we have the following properties:

$$(1)$$ The points $$B,T,C_1,A$$ are on a circle.

$$(2)$$ The points $$B,T,B_1,a$$ are on a circle.

$$(3)$$ The points $$A,B,B_1,S$$ are on a circle.

$$(1')$$ The points $$C,T,B_1,A$$ are on a circle.

$$(2')$$ The points $$C,T,C_1,a$$ are on a circle.

$$(3')$$ The points $$A,C,C_1,S$$ are on a circle.

$$(4)$$ The triangles $$\Delta ABB_1$$, $$\Delta ACC_1$$ are similar.

$$(5)$$ The triangles $$\Delta ABC$$, $$\Delta AB_1C_1$$ are similar.

Before we proceed, we consider the situation obtained by applying an inversion $$X\to X'$$ centered in $$B$$. Let $$D$$ be the intersection of $$BT$$ with the circle $$(T)$$, so that $$BD$$ is a diameter of $$(T)$$. By inversion, circles through $$B$$ are mapped into projective lines, so that $$B\to B'=\infty$$, and $$\infty\to\infty'=B$$. A line through $$B$$ is mapped in the same line. The inversion picture is then constructed using points from the following lemma.

Lemma: Let $$B$$ be a point in the plane. Consider two perpendicular lines passing through $$B$$, and two points $$A', T'$$ on them respectively, so that we have $$BA'\perp BT'\ .$$

Let $$D'$$ be the mid point of segment $$BT'$$.

The perpendiculars in $$D'$$ and $$A'$$ on the lines $$BD'T'$$ and respectively $$BA'$$ intersect in a point $$C'$$.

Let $$U'$$ be the reflection of $$B$$ w.r.t. $$C'$$. (So $$C'$$ is the mid point of $$BU'$$.)

The circle $$(A'U'T')$$ intersects the lines $$A'C'$$ in $$a'$$, so that $$A'a'$$ is the diameter of this circle.

The line $$A'T'$$ intersects $$C'D'$$ in a point $$B_1'$$.

We consider the heights in triangle $$\Delta A'a'B_1'$$, and let its orthocenter be $$C_1'$$. Let $$S'\in B_1'A'$$ be the foot of the height from $$a'$$, $$a'S'\perp B_1'A'$$. and $$BC'U'$$ in $$S'$$.

The circle $$(BS'T')$$ intersects $$D'C'$$ in two points, $$B_1', C_1'$$, chosen so that $$C_1'$$ is between $$D'$$ and $$C'$$.

Then we have:

• $$C_1'$$ is the intersection of the diagonals in the parallelogram $$A'D'T'C'$$, so it is the mid point of $$D'C'$$, and of $$A'T'$$.
• The points $$B, S', C_1', T', B_1'$$ are on the circle.
• The points $$B,S',C',U'$$ are colinear.

Proof of the lemma: Consider the four rectangles with a vertex in $$C'$$ which are congruent to the rectangle $$A'C'D'B$$, and the other vertices are among $$A',B,D',T',U'$$ or reflections of them w.r.t. $$C'$$. We see that $$A'C'$$ goes through the mid point of $$T'U'$$. So $$A'C'$$ is the side bisector of $$T'U'$$ (in the isosceles triangle $$\Delta A'T'U'$$), this implies that $$A'a'$$ is a diameter in the circumcircle $$(A'S'T'a'U')$$ of this triangle. So $$A'T'\perp T'a'$$.

By construction, $$C_1'=A'T'\cap B_1'C'$$ (intersection of two heights), so $$C_1'= D'C'\cap A'T'$$ is the intersection of the diagonals in the parallelogram $$A'C'T'D'$$.

The triangles $$\Delta B_1'C_1'T'$$, $$\Delta B_1'C_1'S'$$ have both a right angle opposite to the side $$B_1'C_1'$$, so $$B_1'C_1'$$ is the diameter of the circle $$B_1'T'C_1'S'$$, and on this circle there is also $$B$$, the reflection of $$T'$$ w.r.t. the same diameter.

We need now to show that $$B',S',C'$$ are colinear. For this we compute the angle \begin{aligned} \widehat{BS'C'} &= \widehat{BS'B_1'} +\widehat{B_1'S'C_1'} +\widehat{C_1'S'C'} \\ &= \widehat{BT'B_1'} +90^\circ +\widehat{C_1'A'C'} \\ &= \widehat{A'a'T'} +90^\circ +\widehat{T'A'C'} \\ &= 90^\circ +90^\circ =180^\circ\ . \end{aligned} This finishes the lemma.

$$\square$$

Proof of the proposition: Lines of colinearity containing points $$X',Y',Z',\dots$$ in the lemma correspond to circles through $$B$$ containing points inverted points $$X,Y,Z$$. So we have the claimed propositions $$(1)$$, $$(2)$$, $$(3)$$; $$(1')$$, $$(2')$$, $$(3')$$. Let us use these circles to obtain equalities of angles. We have: \begin{aligned} \widehat{ABS} &= \widehat{AB_1S} &&\text{ since (ABB_1S) cyclic.}\\ \widehat{ACS} &= \widehat{AC_1S} &&\text{ since (ACC_1S) cyclic. Passing to suplements:}\\ \widehat{ACB} &= \widehat{AC_1B_1}\ . \\ \widehat{CAC_1} &= \widehat{CSC_1} &&\text{ since (ACC_1S) cyclic,}\\ &= \widehat{BSB_1} \\ &= \widehat{BAB_1} &&\text{ since (ABB_1S) cyclic. This implies}\\ \widehat{BAC} &= \widehat{B_1AC_1} \ . \end{aligned}

$$\square$$