Ceva's Theorem

Ceva’s theorem is a theorem regarding triangles in Euclidean Plane Geometry. Consider a triangle ABC. Let CE, BG and AF be a cevians that forms a concurrent point i.e. D. It is stated as follows:

Ceva’s Theorem

Then according to Ceva’s theorem,

\(\large\frac{AG}{GC} \times \frac{CF}{FB}\times\frac{BE}{EA}=1\)

Also, the converse of the above is true, i.e. if \(\large\frac{AG}{GC} \times \frac{CF}{FB}\times\frac{BE}{EA}=1\), then the line AF, BG, CE forms a concurrent point at D.

Let h1 and h2 be the altitudes and the area of a triangle be ABG. When h1 and h2 are constructed the BGC is equal to 0.1(GC)(h1) and ABG is equal to 0.5(AG)(h1). DGC is equal to 0.5 (GC)(h2) and ADG is equal to 0.5 (AG)(h2).

\(\frac{AG}{GC}\frac{(0.5)}{(0.5)}\frac{(h1)}{(h1)}\frac{(AG)}{(GC)}=\frac{(ABG)}{(BGC)}\)

and

\(\frac{AG}{GC}\frac{(0.5)}{(0.5)}\frac{(h2)}{(h2)}\frac{(AG)}{(GC)}=\frac{(ADG)}{(DGC)}\)

Then we have,

\(\frac{AG}{GC}=\frac{(ABG)}{(BGC)}=\frac{ADG}{DGC}\)

\(\frac{AG}{GC}=\frac{(ABG)}{(BGC)}=\frac{ADG}{DGC}\)

Concluding we have,

\(\frac{AG}{GC}=\frac{BDA}{BDC}\)

The resulting equation yields,

\(\frac{CF}{FB}=\frac{ADC}{BDA}\)

And

\(\frac{BE}{EA}=\frac{BDC}{ADC}\)

Thus we have,

\(\frac{(AG)}{(GC)}\frac{(CF)}{(AB)}\frac{(BE)}{(EA)}=\frac{(BDA)}{(BDC)}\frac{(ADC)}{(BDA)}\frac{(BDC)}{(ADC)}\)

By cancellation the resulting equation yields,

\(\frac{(AG)}{(GC)}\frac{(CF)}{(AB)}\frac{(BE)}{(EA)}=1\)

Ceva’s Theorem – Converse

Ceva’s Theorem

We have,

\(\frac{(AG)}{(GC)}\frac{(CF)}{(AB)}\frac{(BE)}{(EA)}=1\)

Here CE, BG, and AF Cevians are concurrent.

Estimate that Cevians CE and AF intersect at D and assume that the Cevians passing through D is BH. So according to Cevians Theorem we have,

\(\frac{AH}{HC}\frac{CF}{FB}\frac{BE}{EA}=1\)

As assumed

\(\frac{AG}{GC}\frac{CF}{FB}\frac{BE}{EA}=1\)

According to Transitive Property, we have

\(\frac{AH}{HC}\frac{CF}{FB}\frac{BE}{EA}=\frac{AG}{GC}\frac{CF}{FB}\frac{BE}{EA}\)

By simplifying

\(\frac{AH}{HC}=\frac{AG}{GC}\)<

It holds true when H and G illustrate the same point. Hence BG, CE, and AF should be concurrent.  


Practise This Question

What are the mid-points of the class-intervals called?