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In the below drawing, we are given: Triangle ABC is right. D is the midpoint of AB. Angle $ACD = φ$ and $BCE = φ$. Angle $EAB = φ$. $ED=α$ and $CD=4a$.

We are looking for angle φ (to be solved by using Geometry, not trigonometry).

What I have tried so far:

Since angles EAB and ECB are both φ, and AB is vertical to BC, then their other sides must also be vertical, so AEC is right.

Therefore the quadrilateral AEBC is inscribed in a semicircle (which I have drawn to see if I get any clue out of it) and AC is its diameter.

If G is the center of the circle, then EG and BG are equal to the circle radius. Also angle $BDC = φ + BAC = EAC$ so triangles EAC and BDC are similar.

But I don't know how to use $ED=a$ and how to combine all this, to calculate angle φ. By the way, I have found it in Geogebra to be ~19 degrees.

Marius Stephant
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1 Answers1

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Alright, I constructed the geometry and verified that the angle is 19 degrees, and I can't tell you how to calculate this angle without trigonometry, but I can tell you how this structure is built, and what the lengths $ \alpha $ and $ 4 \alpha $ reveal. Buckle up!

$$ \widehat {ACE} = \widehat {DCB} $$ Therefore the right triangles $ ACE $ and $ DCB $ are similar. As a result: $$ \frac{AC}{DC} = \frac{AE}{DB} \qquad(1)$$ Now recall that $ CD $ is the median of $ AB $, so $$ AD = DB $$ And replacing in (1) we will have: $$ \frac{AC}{DC} = \frac{AE}{AD} \qquad(2)$$ Now consider the triangles $ACD$ and $EAD$. They have an equal angle ($\varphi$) and equal proportions of the lengths of the sides of that angle, as shown in (2). Therefore these two triangles are similar. And here is an interesting result: $$ \frac{DC}{AD} = \frac{AD}{DE} $$ Or $$ AD^2 = DC.DE $$ Now, recalling that $DE = \alpha$ and $DC = 4\alpha = 4DE$ we have: $$ AD^2 = 4DE^2 $$ $$ AD = 2DE \qquad(3) $$ Et Voila! $$ DC = 4DE = 2AD = 2DB \qquad(4) $$ Look at the right triangle $DBC$ : side $DB$ is half the hypothenuse $CD$ , which means angle $\widehat{DCB}$ must be 30 degrees.

So, to construct the angle $\varphi$ , all we need to do is draw the right triangle $DBC$ , extend the side $BD$ by equal length from point $D$ to find point $A$ , and connect $A$ to $C$ . The angle $\widehat{ACD}$ will be our $\varphi$ .

Saeed
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  • Very nice and thank you for all your work! But from this, how do we deduce that φ = 19? – Marius Stephant Dec 08 '20 at 07:32
  • OK got it!! $AE = α\sqrt{7}$. So now in triangle AED we have all 3 sides: $α, 2α$ and $α\sqrt{7}$. – Marius Stephant Dec 08 '20 at 07:59
  • Thanks. You are right about the angle: I can draw it but can't calculate it. And you are right about sizes of triangle AED. I should have added this in the text: Starting with triangle DBC, its sides are $2\alpha$ , $\frac{\sqrt 3}{2}\alpha$ and $4\alpha$ , we can go on to triangle ABC, whose hight is $4\alpha$ and hypotenuse $2\alpha\sqrt{7}$ , and then note that because angle $\widehat{CAE}$ in right triangle AEC is 60 degrees, we have $EG=GA=AE$ , meaning AE equals the radius of the big circle, or half the hypotenuse of ABC, hence $\alpha\sqrt7$ . – Saeed Dec 08 '20 at 13:10
  • The exact value of the $\phi$ is $19.10660535086909439451747474013^o$. Determining its value is not possible without resorting to trigonometry. – YNK Dec 08 '20 at 16:03