# Proving a Line to bisect a line in a Triangle

From point $A$ tangents $AB$ and $AC$ to a circle are drawn ($B$ and $C$ tangent points); $PQ$ is a diameter of the circle; line $L$ is tangent to the circle at point $Q$. Lines $PA$, $PB$, and $PC$ intersect line $L$ at points $A_1, B_1, C_1$. Prove that $A_1B_1 = A_1C_1$.

• So, what have you tried?
– amd
May 3 '17 at 1:40
• i was thinking of trying to prove $\triangle PA_1B_1 \equiv$ $\triangle PA_1C_1$ by AAS or ASA...but i havent been successful in that May 3 '17 at 1:40
• @rover2 those triangles won't be equivalent in general. In your diagram, if A is moved up $\angle B_1PA_1$ gets smaller and $\angle C_1PA_1$ gets larger, in at least one case - which I drew myself.
– Χpẘ
May 3 '17 at 19:46
• there was one hint which was provided to me on here in which it said to extend $PQ$ until it intersects the tangents at points $B$ and $C$ at point $A$. using this method it just suffices to show the two triangles are similar. the diagram i included in my question was a diagram i came up with based on what the hypothesis provided @Χpẘ May 4 '17 at 1:56
• @rover2 The incircle of the triangle formed by the lines $AB, AC$ and $B_1C_1$ is tangent to line $B_1C_1$ at point $A_1$. Everything else follows from this fact. May 4 '17 at 11:18

My idea is to show that by taking $M$ as the midpoint of $B_1C_1$, we prove that $BA, AC$ and $PM$ meet at the same point $A$ and hence $M=A_1$. To do this, we can look at the triangle $BP_1C_1$. (For this paragraph, refer to the original image of the problem statement, thanks.)

We do a change of variables to make it clearer, and draw a image to make it more understandable. (Note: the circle in the problem is the circumcircle of $\triangle BDF$, while the $FG$ line in my graph refers to the tangent line $BA$ in the problem.)

Let us discuss an acute triangle $\triangle ABC$ with $BD\perp AC$ at $D$, $DF\perp AB$ at F, $E,M$ being the midpoints of $BD$ and $AC$ respectively, and $FG_A\perp EF$ such that $FG_A$ intersects $BM$ at $G_A$. We do the same operation on the other side of $BD$ to get $G_C$. Our goal is to prove $BG_A=BG_C$ and conclude that $GB$ bisects $AC$.

By the Sine Law, $$BG_A = \sin \angle BFG_A\cdot \frac{BF}{\sin\angle FG_AB}=\sin A\cdot\frac{BF}{\sin(A-\angle ABM)} = BD\cdot\frac{\sin^2 A}{\sin (A-\angle ABM)}$$ and similarly we can get $$BG_C = BD\cdot\frac{\sin^2 C}{\sin (C-\angle CBM)}$$ Hence, our goal is to prove $BG_A=BG_C$ which will give us $G_A=G_C$: $$\frac{\sin^2 A}{\sin (A-\angle ABM)}=\frac{\sin^2 C}{\sin (C-\angle CBM)}$$

Let us tackle this by rewriting the equation to $$\frac{\sin A}{\cos\angle ABM-\cot A\sin \angle ABM}=\frac{\sin C}{\cos\angle CBM-\cot C\sin \angle CBM}$$

Look at the image below. We let $MT\perp AB$ at $T$, and $MK \perp BC$ at $K$. Then, the above equation can be rewritten as $$\frac{\frac{BD}{AB}}{\frac{BT}{BM}-\frac{AD}{BD}\cdot\frac{MT}{BM}}=\frac{\frac{BD}{BC}}{\frac{BK}{BM}-\frac{CD}{BD}\cdot\frac{MK}{BM}}$$ which is equivalent to proving $$AB\cdot BD\cdot BT - AB\cdot AD \cdot MT = BC\cdot BK \cdot BD- BC\cdot CD \cdot MK$$

We can simplify this by noticing that $\triangle AMT\sim \triangle ABD$ so $AB\cdot MT=AM\cdot BD$, and $\triangle CMK\sim \triangle CBD$ so $BC\cdot MK=CM\cdot BD$ Hence, proving the equation is equivalent to proving $$AB\cdot BT - AM\cdot AD = BC\cdot BK - MC\cdot CD$$

Notice that $AM\cdot AD = AT\cdot AB$ and $CM \cdot CD = BC \cdot CK$, so we only need to prove $$AB\cdot (BT-AT) = BC \cdot (BK-CK)$$ which is proved by noticing that $ASJC$ is cyclic, with $AT=ST$ and $CK = JK$, so $BS\cdot BA = BJ \cdot BC$.

Therefore, $G_A=G_B=G$ is proved. Now we may return to our original problem, and see that $G$ corresponds to point $A$, while the points $(ABCDEFM)$ correspond to the points $(B_1PC_1QOBA_1)$ respectively, and hence $PA$ must bisect $B_1C_1$ as desired.

• Best acceptable answer . May 3 '17 at 4:18
• Thanks, the solution is a bit messy and may have some unclear parts, so if you spot any problems please let me know. May 3 '17 at 4:26

Not a complete solution yet, but this is the beginning of pure "bashing" with coordinates.

Choose a coordinate system so that $B_1 = (1,0)$ and $C_1 = (0, -1).$ Then, it suffices to prove that $A_1 = (0,0).$

Let $P = (a,2b)$ and $Q = (a,0)$, so the equation of the circle is $(x-a)^2+(y-b)^2 = b^2$. This is simplified to $x^2+y^2-2ax-2by+a^2 = 0.$ The equation of the line passing through the points $P$ and $B_1$ is simply $y=\dfrac{2b}{a-1}x-\dfrac{2b}{a-1}$. Thus, coordinate of the point $B$ will satisfy the above two equations, since it is the intersection of the circle and that line. Substituting $y$ in terms of $x$ and denote $p = \dfrac{2b}{a-1}$, we get: $$0 = (1+p^2)x^2-2x(a+bp+p^2)+(p^2+2bp+a^2).$$

Now when you solve this equation, you will get two solutions $x_1 = a$ and $x_2 = \dfrac{p^2+a}{p^2+1}$, but we have to take the latter because $P\neq B.$ Thus, $B = \Big(\dfrac{p^2+a}{p^2+1}, \dfrac{2b}{p^2+1}\Big)$.

Similarly, the equation of the line passing through $P$ and $C$ is $y = \dfrac{2b+1}{a}x - 1$ and when you plug this into the equation of the circle you will find $C = \Big(\dfrac{q+a}{q^2+1}, \dfrac{aq-1}{q^2+1} \Big)$, where $q = \dfrac{2b+1}{a}.$

Now, it is straightforward to find the coordinate of point $A$ because it is the intersection of two tangents to the circle at $B$ and $C$. After this, find the equation of the line $l = PA$, then the coordinate of $A_1$ will be found by intersecting $l$ with the $x$ axis.

I will not have time to complete the remaining computations, but it is completely straightforward at this point. But our goal is to prove that $A_1 = (0,0)$; therefore, if the problem is true, then it the equation of the line $l$ must have been found in the form $y = cx, c\in\mathbb{R}.$

Let the circle from the problem is denoted by $k$ and let $O$ be its center. Let $M = AB \cap B_1C_1$ and $N = AC \cap B_1C_1$. Through point $A_1$ draw a line parallel to $PQ$ and let that line intersect $AO$ at point $I$. Define circle $k_I$ with center $I$ and radius $A_1I$.

Lemma 1. Circle $k_I$ is the incircle of triangle $AMN$.

Proof: By assumption, line $B_1C_1$ is tangent to circle $k$ at point $Q$ and $PQ$ is a diameter of $k$. Therefore $PQ \perp B_1C_1$. By construction, $A_1I || PQ$, hence $A_1I \perp B_1C_1$. Then, circle $k_I$ is tangent to $B_1C_1$ as well.

Furthermore, let us perform a homothety with center $A$ which maps point $P$ to point $A_1$. Since the image of line $PQ$ under the homothety should be the line through $A_1$ parallel to $PQ$ and since by construction $A_1I || PQ$, the image of line $PQ$ under the homothety is line $A_1I$.

Furthermore, line $AO$ is mapped to itself, thus the intersection point $O = PQ \cap AO$ is mapped to the intersection point $I = A_1I \cap AO$. Consequently, segment $PO$ is mapped to segment $A_1I$. However, $PO$ is the radius of circle $k$ and $O$ is its center, while $A_1I$ is the radius of circle $k_I$ and $I$ is its center. Therefore, the circle $k$ is mapped by the homothety to circle $k_I$. Since $k$ is tangent to both lines $AB$ and $AC$, which are mapped to themselves by the homothety, the circle $k_I$ is also tangent to $AB$ and $AC$. Consequently, $k_I$ is the incircle of triangle $AMN$.



Lemma 2. $MQ = MB_1 = MB$ and $NQ = NC_1 = NC$.

Proof: Since $PQ$ is a diameter of circle $k$, $\, \angle \, PBQ = 90^{\circ}$ which means that $QB \perp PB_1$ and therefore $\angle \, QBB_1 = 90^{\circ}$. Since $MB$ and $MQ$ are tangents to $k$, $\, QM = BM$ which in the right-angled triangle $QBB_1$ implies that $MQ = MB_1 = MB$. The proof of the second chain of identities is analogous.



Lemma 3. $MA_1 = NQ = NC_1 = NC$ and $NA_1 = MQ = MB_1 = MB$.

Proof: Denote by $$p = \frac{AM + MN + AN}{2}$$ the half perimeter of triangle $AMN$. As $k_I$ is the incircle of $AMN$ by lemma 1, $\, A_1$ is the point of tangency of the incircle $k_I$ of triangle $AMN$ with the edge $MN$, hence $$MA_1 = p - AN$$ Observe that $AB$ and $AC$ are the two tangents to $k$ from $A$ so $AB = AC$. Analogously $MQ$ and $MB$ are the two tangents from $M$ so $MQ = MB$. Analogously $NQ$ and $NB$ are the two tangents from $M$ so $MQ = MB$. Thus $$MN = MQ + NQ = MB + NC$$ so $$2 \, p = AM + MN + AN = AM + (MQ + NQ) +AN = AM + (MB + NC) + AN = (AM + MB) + (AN + NC) = AB + AC$$ Combined with the fact that $AB = AC$ one concludes that $2\, p = AB + AC = 2 \, AC$ so $$AC = p = AB$$. Therefore, by lemma 2 $$NC = AC - AN = p - AN$$ Thus, $$MA_1 = p - AN = NC = NC_1 = NQ$$ The second chain of identity can be proved analogously.



Concluding the proof: By lemma 2, $MB_1 = MQ = MA_1 + QA_1$ so $$A_1B_1 = MA_1 + MB_1 = MA_1 + MA_1 + QA_1 = 2 \, MA_1 + QA_1$$ By lemma 3, $$A_1C_1 = QA_1 + QC_1 = QA_1 + NQ + NC_1 = QA_1 + MA_1 + MA_1 = 2 \, MA_1 + QA_1$$ Therefore, $$A_1B_1 = 2 \, MA_1 + QA_1 = A_1C_1$$

Brute force will do as well. Choose a coordinate system where the circle's center is the origin and $P=(0,-r)$, $r$ being the circle's radius. Let $A=(a_1,a_2)$, $l^2=a_1^2+a_2^2$ and $w=\sqrt{l^2-r^2}$. Now calculate $B$ and $C$ from

$$x^2+y^2=r^2\quad\text{and}\quad a_1x+a_2y=r^2;$$

the latter is the equation of the polar with pole $A$. Hence the $y$-intercepts of the lines $PB$ and $PC$ are $$\frac{a_2r\mp a_1w}{a_1r\pm a_2w+l^2}r.$$ The average of those intercepts equals $$\frac{a_2}{a_1+r}r,$$ which is the $y$-intercept of the line $PA$ by the intercept theorem.

Remark: the calculation takes a full page ... Not esthetical, but pleasing at the end as all cancel off very nicely.