In $\triangle ABC$, $AD$ $\perp$ $BC$ and $GE$ is the extended line of $DG$ where $G$ is centroid. Prove that $GD$ = $\frac{EG}{2}$

Let $$ABC$$ be a triangle and in $$\triangle ABC$$, $$AD$$ $$\perp$$ $$BC$$ and three median lines intersect at point $$G$$ where $$G$$ is the centroid of $$\triangle ABC$$. The extension of $$DG$$ intersects the circumcircle of $$\triangle ABC$$ at point $$E$$. Prove that $$GD = \frac{EG}{2}$$

I found this as an isolated problem.

My attempt:

Nothing speciality I discovered from the diagram. I only connected segment $$AE$$ and drew $$GI$$, where $$GI$$ $$\perp$$ $$AD$$.

From the above diagra, $$G$$ is the centroid. So, $$\frac{AG}{GF}$$ = $$\frac{1}{2}$$. And then from right angled triangle $$\triangle AGI$$ and $$\triangle ADF$$, We get $$AI:ID$$ = $$1:2$$ (as $$\triangle AGI$$ $$\sim$$ $$\triangle ADF$$).

Right then, if $$\triangle ADE$$ can be showed as a right angled triangle ($$\angle EAD$$ = 90$$^\circ$$) and $$\triangle ADE$$ $$\sim$$ $$\triangle GID$$, we can also likewise show that $$\frac{DG}{GE}$$ = $$\frac{1}{2}$$.

But my reverse effort went into vain. I can't anyhow show that $$\angle EAD$$ = 90$$^\circ$$. So, how to solve for that case?

Can it be solved by vector? Thanks in advance.

• The answer you accepted is deficient. – Aqua Feb 16 '19 at 0:27

What you only have to prove is that $$[CE]\parallel [AB]$$. From there it's trivial (observe the ratio $$CG:GF$$) that $$[GD]=\frac{[EG]}{2}$$

So here goes my proof for $$[CE]\parallel [AB]$$ (I guess there might be a simpler one, however...)

Here $$FK$$ is the perpendicular bisector of $$[AB]$$, $$K=DE\cap FK$$ and $$I$$ is the midpoint of $$[CD]$$. Now since $$CD\parallel KF$$ $$\frac{[CD]}{[FK]}=\frac{[CG]}{[GF]}\implies [CD]=2·[FK]$$ Therefore $$[KI]\parallel [AB]$$, which implies that $$[CK]=[DK]$$.

Denote by $$O$$ the circumcenter.

Simple angle-chasing shows that $$\angle CKO=\angle OKE$$. From the congruence criterion SAS we obtain $$\Delta OKC\cong \Delta OEK\implies \angle KOC=\angle EOK$$
Thus the triangle $$\Delta OEC$$ is isosceles; the angle bisector of $$\angle EOC$$ is $$OK\perp CE$$. We can now conclude that $$CE\parallel AB$$

• @Doctor I didn't understand a fact. How can point $I$ be the midpoint $CD$? If you please explain that, it will be too much better for me. – Anirban Niloy Feb 15 '19 at 17:16
• It's not a conclusion but the designation. It's just like saying 'I will denote with $I$ the midpoint of $[CD]$ – Dr. Mathva Feb 15 '19 at 17:22
• And sorry for changing the letters of the vertices. I noticed that when the answer was finish... – Dr. Mathva Feb 15 '19 at 17:23
• No problem. Main fact is that I understood the total process. I have to say you tnx cordially for a commendable approach. And there is no reason for being sorry. – Anirban Niloy Feb 15 '19 at 17:26
• Why is that: $[KE]$ is the reflection of $[CK]$ over $KF$? This is true if $AB||CE$ but that is the essence of the prove. – Aqua Feb 16 '19 at 0:22

Before solving the problem, let's put ourselves in this situation (see attached image)

If $$HD=DJ, HG = 2(GO)$$ and $$JO=OI$$ then you have to necessarily $$J, O$$ and $$I$$ are collinear.

A simple test is by contradiction. Suppose they are not collinear, so let's place $$L$$ in $$HO$$ such that $$DL // JO$$ then we would have $$HL = 3a$$, $$LG = a$$ and $$GO = 2a$$ ($$a$$ is a constant), in addition $$DL = 2 (JO) = 2 (OI)$$. With all this, we would have that $$DL // OI$$ (the triangles $$DLG$$ and $$GOI$$ are similar). Let's call $$\theta = \angle JOL$$, then $$\angle DLH = \theta$$ and $$\angle DLO = \angle LOI = 180 - \theta$$, therefore $$J,O,I$$ are collinear.

In the problem, let's call $$H$$ the orthocenter of the triangle $$ABC$$ and $$O$$ to its circumcenter. It is known that $$H,G,O$$ are collinear and $$HG = 2GO$$. Now let $$J$$ be the intersection of the extension of $$AH$$ with the circumscribed circumference, so it is easy to see that $$HD = DJ$$. Finally, we would have by the initial observation that necessarily J, O, I are collinear. Then $$G$$ is the centroid of the triangle $$HIJ$$ and $$GI=2DG$$.

Sorry for having changed the letter $$E$$ for the $$I$$, I hope it is understood in the same way

Let $$E'$$ be such that $$ABCE'$$ is isosceles trapezoid ($$AB = CE'$$).

Then $$E'$$ is on circle through $$A,B,C$$. Let $$DE'$$ meet $$AF$$ at $$G'$$.

Since triangle $$DFG'\sim E'AG'$$ and $$AE' = 2DF$$ we have $${AG'\over G'F} = {E'G'\over G'D} = {2\over 1}$$ so $$G'=G$$ and $$E'=E$$ and the claim is proved.

• You provided me with an efficient solution. But without diagram, my brain doesn't work. Moreover, my desired condition that I have asked for is to show the $CE$ $\parallel$ $AB$. Any idea? Then please add that to your post if Dr. Mathva has done mistake. – Anirban Niloy Feb 16 '19 at 3:25
• I thought you accept my answer? – Aqua Feb 24 '19 at 10:38