# Show that if $\angle ADB = 60^{\circ}$ then $AA_1 = BB_1$ (and answer whether the converse is true).

In the diagram above we have that $AA_1$ and $BB_1$ are altitudes and $\angle ADB = 60^{\circ}$. The problem is two fold- show that from$\angle ADB = 60^{\circ}$ it follows that $AA_1$ = $BB_1$ and secondly answer whether it is true that $AA_1=BB_1 \implies \angle ADB = 60^{\circ}$.

Here is my proof for the first part: Since the sum of the angles in a triangle add up to $180^{\circ}$ we have that $\angle DAA_1 = DBB_1 = 30^{\circ}$. Furthermore, it can be proven that the orthocenter $H$ and $B_1$ (also $A_1$) are symmetric about $AD,DB$ respectively (let me know if you want me to add this proof, I think the question would get too convoluted). Therefore the triangles $AHB_1$ and $BHA_1$ are equilateral from which the theorem follows.

I have trouble with the second part as I think the statement is true but in my book it says it is false. Here is my proof and the question is where is my mistake.

Assume that $AA_1=BB_1$. It can be proven (let me know for this also) that if two chords in a circle are equal they form isosceles triangles from the point of intersection. Thus $HB_1=HA$. But we also have that $H$ and $B_1$ are symmetric about $AD$, so $HA = AB_1$ so $\angle B_1AH = 60^{\circ}$ and furthermore $AD$ is its bisector so $DAH = 30^{\circ}$ and from $180^{\circ}$ theorem about triangles it follows that $\angle ADB = 60^{\circ}$.

Again,my question is what is the problem with the second proof.

• Are you sure that $HB_1=HA$? Couldn't it be that $HA=HB$? – RicardoCruz Dec 9 '15 at 13:58
• Yes I am completly sure about this theorem. Should I post a proof of it ? – alexgiorev Dec 9 '15 at 14:07
• For $HA=HB$ and $HB_1=HA_1$ you got also isosceles triangles from the point of intersection. – RicardoCruz Dec 9 '15 at 14:10
• I have misspelled the theorem. Let me try to type it correctly (but $HB_1 = HA$ definetly holds) – alexgiorev Dec 9 '15 at 14:12
• Look, you know that two chords which are not trough the center cannot bisect each other right? So they are split into unequal segments and the theorem is that the smaller segment from one chord is equal to the smaller from the other. Is that more clear? We cant have that $HA = HB$ as that would make the triangle isosceles and that is not part of the assumptions. – alexgiorev Dec 9 '15 at 14:14

I am playing your suggestion using Geobebra. From my construction, $BB_1 = AA_1’ = AA_1$, but the shaded triangles drawn clearly are not isosceles. Let me explain how it is constructed so that you can play with it too:-

1. Draw a circle (in red) and select a chord $BB_1$.

2. On the circle, select point $A$ (other than $B$ or $B_1$).

3. After creating the vector $BA$, translate $BB_1$ to $AA_1’$ such that $BB_1 = AA_1’$.

4. Let the dotted circle (centered at $A$, radius $= AA_1’$) cut the red at $A_1$ (an intersection point).

5. Let $D$ be the intersection point of $AA_1$ and $BB_1$.

6. Moving point $B_1$ around, you will find that on some occasions, (1) the two triangles are not isosceles but (2) sometime they are.

It is not difficult to see that (2) is true only when $A_1B(A_1’)$ is parallel to $AB_1$.

• You are correct that the shaded are not isosceles but consider the ones with base B1A1 and the other has base AB. The problem with my proof is that the position of the altitudes in my diagram is like so only when the triangle is acute. – alexgiorev Dec 10 '15 at 6:07
• Right. I forgot all about the altitudes. I will try to fix that...later. – Mick Dec 10 '15 at 7:22