# Finding angle in a triangle with one median and an isosceles triangle in it

In the following figure $$AF=BD=DC$$ and $$AE=EF$$. Find the angle $$\alpha$$.

I tried so many different things (constructing triangles, drawing parallel lines, using Ceva theorem in triangles $$ABD$$ and $$CBE$$, ...). Any help for solving this will be much appreciated.

A geomtrical solution (without trigonometry) would be very nice.

• @cosmo5 is it plausible, in your mind, that there is no typo? Mar 14, 2021 at 15:30
• Its not a typo. Mar 14, 2021 at 15:44
• @cosmo5 The answer is exactly 60 degrees according to Geogebra. Mar 14, 2021 at 15:46
• @MathLover I made same mistake as you. It is AF=BD=DC not AD... Mar 14, 2021 at 15:50
• @Ghartal No, for $AF=BD=DC$ Mar 14, 2021 at 17:09

In isosceles $$\triangle AEF$$, drop $$EG \perp AF$$ with $$G$$ on $$AF$$. Let $$AG=GF=x$$. So $$BD=2x$$. Let $$\angle BAD = \theta$$. So $$AE=x\sec \theta$$.

Applying Menelaus' for transversal $$CE$$ to $$\triangle ABD$$, $$\frac{BC}{DC}\cdot\frac{DF}{FA}\cdot\frac{AE}{EB}=1$$ $$\Rightarrow DF=EB\cos \theta$$ $$\Rightarrow AD-2x=(AB-x\sec\theta)\cos \theta$$ $$\Rightarrow AD=AB\cos \theta+x$$

Next drop $$BH \perp AD$$ with $$H$$ on $$AD$$. In right $$ABH$$, $$AH=AB \cos \theta$$. Then $$AD=AH+DH$$ implies $$DH=x$$!

Hence in right triangle $$BHD$$, $$BD=2x$$, $$HD=x=BD/2$$. It turns out $$\triangle BHD$$ is $$30^{\circ}-60^{\circ}-90^{\circ}$$ and we conclude $$\alpha = 60^{\circ}$$

• Cool trigonometric proof. There is also a geometric proof, if you are curios. Mar 20, 2021 at 19:02

Here is a purely geometric proof without any trigonometry.

Draw the three lines $$a, \, b, \, c$$ such that: $$A \, \in \, a \,\,\text{ and } \,\, a \, || \, BC$$ $$B \, \in \, b \,\,\text{ and } \,\, b \, || \, CA$$ $$C \, \in \, c \,\,\text{ and } \,\, c \, || \, AB$$ Furthermore, let $$B^* \, = \, a \, \cap \, c \,\,\, \text{ and } \,\,\, A^* = b \, \cap \, c$$ Then, $$ABA^*C$$ and $$ABCB^*$$ are parallelograms and furthermore $$AB = CA^* = CB^*$$ as well as $$AB^* = BC$$ Apply Menelaus' theorem to the triangle $$BCE$$ intersected by the line that passes through the collinear points $$A, \, F,\, D$$: $$\frac{AB}{AE} \cdot \frac{EF}{CF} \cdot \frac{DC}{BD} = 1$$ Since by assumption $$AE = EF$$ and $$BD = CD$$, one concludes that $$AB = CF$$ Combining this latter fact with the earlier conclusions, one observes that $$CF = AB = CA^* = CB^*$$ which is possible if and only if triangle $$\Delta \, A^*B^*F$$ has $$90^{\circ}$$ angle at vertex $$F$$, i.e. $$\angle \, A^*FB^* = 90^{\circ}$$. But that means that: $$\angle \, AFB^* = 180^{\circ} - \angle \, A^*FB^* = 180^{\circ} - 90^{\circ} = 90^{\circ}$$ Recall that by assumption $$AF = BD = DC = \frac{1}{2}\,BC$$. But by construction, $$ABCB^*$$ is a parallelogram and $$AB^* = BC$$ so $$\Delta \, AFB^* \, \text{ is a triangle with the properties that } \, \angle \, AFB^* = 90^{\circ} \,\, \text{ and } \,\, AF = \frac{1}{2}\, AB^*$$
which is possible if and only if $$\angle \, FAB^* = 60^{\circ}$$ However, $$AB^*$$ is parallel to $$BC$$ and consequently, by the properties of a pair of parallel lines intersected by a third line, $$\angle \, ADB = \angle \, DAB^* = \angle \, FAB^* = 60^{\circ}$$

• You are right, I got it now! I neglected the collinear requirement and you did give a beautiful answer. I will upvote yours and delete my answer. Thank you for the challenge. Mar 20, 2021 at 19:21
• Nice one. You also find a $30-60-90$ triangle using construction. Mar 21, 2021 at 5:18