3
$\begingroup$

Given is a quadrilateral $ABCD$ in which $\angle DAB=\angle CDA=90$. Point M is the midpoint of side $BC$ and circumscribed circles of triangles $\triangle ABM$ and $\triangle DCM$ meet at points $M$ and $E$. The line $EC$ intersects the circumscribed circle of $ABM$ second time in $F$, and lines $AF$ and $CD$ meet at $G$. Prove $EB=EC$ and that $F,M,G,C$ are concyclic.

Image

I've only worked on $EB=EC$ as I assume that's a prerequisite to $F,M,G,C$ being proven to be concyclic. I think we might have to prove that $E$ has to be on line $AD$ as we know that since $\triangle EBC$ should be an isosceles triangle and $M$ is midpoint of $BC$ the angles $\angle EMC=\angle EMB=90$, but also we know that since $CDEM$ is an cyclic quadrilateral that $\angle EMC+\angle EDC=180$ and $\angle EDC=\angle CDA-\angle ADE$ so $\angle ADE=0$? Any help's appreciated, thanks!

$\endgroup$

3 Answers 3

Reset to default
2
$\begingroup$

enter image description here

Hint:

Prove that $\angle AFM= \angle DCM$ by considering their relationships with $\angle ABC$.

$\endgroup$
6
  • $\begingroup$ Thanks! I've figured this part out, how to go about proving $EB=EC$ $\endgroup$ Commented Dec 18, 2022 at 11:36
  • $\begingroup$ Join $EM$. Prove that $\angle EMC=90^o$ and then $\Delta EMC \cong \Delta EMB$. $\endgroup$ Commented Dec 18, 2022 at 11:55
  • $\begingroup$ I'm aware of that, I mentioned it in my original post above, but isn't a prerequisite of that for $E$ to be on $AD$? How would one prove that? Can you elaborate? $\endgroup$ Commented Dec 18, 2022 at 11:57
  • $\begingroup$ (1) We can prove that $A, D, E$ are collinear by noting that $\angle AEM=180^o-\angle ABC$ and $\angle DEM=180^o-\angle BCD$. (2) But I don't see why we need to prove this? $\endgroup$ Commented Dec 18, 2022 at 12:07
  • $\begingroup$ Oh, I see, thanks! We need to prove this because if $BCE$ is an isosceles triangle then it's height from $E$ would also be its' median and therefore $\angle EMC=90$ which means, from the fact that $EMCD$ is a cyclic quadrilateral that $\angle EDC+\angle EMC=180$ so $\angle EDC=90$ which is only possible if $A,D,E$ are colinear (because we know from the statement that $\angle ADC=90$) $\endgroup$ Commented Dec 18, 2022 at 12:10
2
$\begingroup$

If you proved $EB = EC$ the second one follows easily from there (couldn't comment cause of low rep.). If $EB=EC$ notice that $E$ has to lie on $AD$ since we have $\angle EMC= \angle CDA=90$ (basically EC and EB will be the diameters of the circles).
To prove points F, M, C, G are concyclic you can simply assign few angles to see it. Lets say $\angle EAF = \alpha, \angle EFA = \beta, \angle FCM = \theta$. Since $\angle DEC = \alpha + \beta$ and $\angle CEM = 90 - \theta$ angle $\angle DCM = 90 - \alpha - \beta + \theta$. Notice that angle $\angle GFM = 90 - \theta + \alpha + \beta$. Hence, those points are concyclic.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer! The second part of the question seems pretty easy, it looks like the really hard part is proving $EB=EC$ $\endgroup$ Commented Dec 18, 2022 at 11:36
  • $\begingroup$ You mentioned you worked on EB=EC, I thought you got that part. The first part is also straightforward as mentioned in above comment. Once you see E in on AD, EM becomes perpendicular to BC. $\endgroup$ Commented Dec 18, 2022 at 12:58
0
$\begingroup$

We will make use of a couple of corollaries to the Inscribed Angle Theorem. One corollary says that a quadrilateral is concyclic iff opposite angles are supplementary. Another is Thales' Theorem, which says that if the diameter of a circle is a side of an inscribed triangle, that triangle is a right triangle whose hypotenuse is the given diameter. Conversely, the hypotenuse of a right triangle is the diameter of its circumcircle.

Let $O_1$ be the circumcenter of $\triangle ABM$ and $O_2$ be the circumcenter of $\triangle MCD$.

Let $E$ be the intersection of the circumcircles of $\triangle ABM$ and $\triangle MCD$.

enter image description here

Since $A,B,M,E$ are concyclic, $$ \angle AEM+\angle ABM=\pi\tag1 $$ Since $C,M,E,D$ are concyclic, $$ \angle MED+\angle MCD=\pi\tag2 $$ Since $\angle ABM+\angle MCD=\pi$, $(1)$ and $(2)$ give $$ \angle AED=\angle AEM+\angle MED=\pi\tag3 $$ That is, $A,E,D$ are colinear.

Since $\triangle EAB$ is a right triangle, $O_1$ is the midpoint of diameter $\overline{EB}$.
Since $\triangle EDC$ is a right triangle, $O_2$ is the midpoint of diameter $\overline{EC}$.

enter image description here

Since $\overline{O_1B}=\frac12\overline{EB}$ and $\overline{BM}=\frac12\overline{BC}$ and $\angle O_1BM=\angle EBC$, SAS similarity says $$ \overline{O_1M}=\frac12\overline{EC}\tag4 $$ Since $\overline{O_1M}=\overline{O_1B}=\frac12\overline{EB}$, $(4)$ implies $$ \overline{EB}=\overline{EC}\tag5 $$

enter image description here

Since $A,B,M,F$ are concyclic, $\angle AFM+\angle ABM=\pi$. Furthermore, $\angle AFM+\angle MFG=\pi$. Therefore, $$ \angle ABM=\angle MFG\tag6 $$ Furthermore, $\angle MCG+\angle ABM=\pi$. Therefore, $$ \angle MCG+\angle MFG=\pi\tag7 $$ which says that $M,F,G,C$ are concyclic.

$\endgroup$
1
  • $\begingroup$ I could have put all the lines and points in one image, however, I thought it was a bit too crowded and not as clear. $\endgroup$ Commented Dec 21, 2022 at 17:19

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.