Prove that $\angle FME=90$ (WITHOUT using triangle identity) and show that $ME=MF$

Question

Prove that $\angle FME=90$ ( DO NOT use triangle identity) and show that $ME=MF$ (Triangle $ABC$ is right at $A$ and $AB=AC$ and $M$ is at the middle of $BC$,also $D$ is an arbitrary point on $BC$ and $DE$ and $DF$ are perpendicular to $AB$ and $AC$ respectively.)

I have already solved this using triangle identity technique(triangles $AMF,BME$ ) but don't know other method to solve it! Though was not able to prove $\angle FME=90$!

Note that $\angle BMA=90^{\circ}$. Hence the equality of $AMF$ and $BME$ evidently implies $\angle FME=90^{\circ}$ — polfosol, Sep 16 '16 at 15:22
I'm not convinced of your reason! Also in this problem we are not allowed to use triangle equality! — Hamid Reza Ebrahimi, Sep 16 '16 at 15:25
Would you accept a solution that assigns coordinates to the points and then proves the stated claims (an algebraic one)? — Parcly Taxel, Sep 16 '16 at 15:31
@HamidRezaEbrahimi ...and I refer you to d'Aurizio's answer for what I had hinted at. — Parcly Taxel, Sep 16 '16 at 15:38

score 2 · Accepted Answer · answered Sep 16 '16 at 15:26

2

The four points $F,D,E$ and $A$ are concyclic. Also, since $\angle{AMD}+\angle{DEA}=180^\circ$, the four points $M,D,E$ and $A$ are concyclic.

Therefore, the five points $F,D,E,A$ and $M$ are concyclic.

It follows from this that $$\angle{EFM}=\angle{EAM}=45^\circ\quad\text{and}\quad \angle{FME}=\angle{FDE}=90^\circ$$

answered Sep 16 '16 at 15:26

mathlove

139,939

What do you mean by concyclic? Do you mean lying on the same circle? – Hamid Reza Ebrahimi Sep 16 '16 at 15:32
@HamidRezaEbrahimi: Yes, see, for example, here. – mathlove Sep 16 '16 at 15:33
$\angle EAM=45$ is obvious,but how could you infer that $\angle EFM=45$? – Hamid Reza Ebrahimi Sep 16 '16 at 15:37
@HamidRezaEbrahimi: Since the four points $E,A,F,M$ are concyclic, we have $\angle{EFM}=\angle{EAM}$. Also, $\angle{EAM}=45^\circ$. Hence, $\angle{EFM}=45^\circ$. – mathlove Sep 16 '16 at 15:39
Thank you(+1),how can we find CONCYCLIC points in a geometric figure? – Hamid Reza Ebrahimi Sep 16 '16 at 15:43
@HamidRezaEbrahimi: See the angles. That $F,D,E,A$ are concyclic comes from that $\angle{DFE}=\angle{DAE}$, for example. Also, as I wrote, that $M,D,E,A$ are concyclic comes from that $\angle{AMD}+\angle{DEA}=180^\circ$. I hope this helps. – mathlove Sep 16 '16 at 15:46
I think $\angle{FME}=90^\circ$ is the direct consequence after showing that M is one of the concyclic point. – Mick Sep 16 '16 at 17:36

score 1 · Answer 2 · answered Sep 16 '16 at 15:25

1

Hint One way is to look at the triangle $ABC$ and its $90^{\circ}$ clock-wise rotation around point $M$. Show that $F$ is the image of $E$.

Or more directly: Look at triangles $FMA$ and $EMB$. Anything special about them? That should do it.

answered Sep 16 '16 at 15:25

Futurologist

9,509

score 1 · Answer 3 · answered Sep 16 '16 at 15:36

1

It is quite trivial with coordinate geometry. Assume that $AE=x$ and $AF=y$. Then $$ D=(x,y),\quad B=(x+y,0),\quad C=(0,x+y),\quad M=\left(\frac{x+y}{2},\frac{x+y}{2}\right) $$ and

$$ MF^2 = \left(\frac{x+y}{2}\right)^2+\left(\frac{x-y}{2}\right)^2 = ME^2 = \frac{1}{2} EF^2$$ so $EMF$ is an isosceles right triangle by the Pythagorean theorem.

answered Sep 16 '16 at 15:36

Jack D'Aurizio

353,855

Thank you(+1) , nice and concise. – Hamid Reza Ebrahimi Sep 16 '16 at 15:44
@HamidRezaEbrahimi: you're welcome. – Jack D'Aurizio Sep 16 '16 at 15:53

Prove that $\angle FME=90$ (WITHOUT using triangle identity) and show that $ME=MF$

3 Answers3