# E is the midpoint of a median AD of triangle ABC and BE is produced to meet AC at F .Show that AF=1/3AC

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See diagram.

We join F and D.

In the two triangles BDF and CDF, wee see that BD = CD. The altitude from F on to BC will be same for both triangles. Hence Area of triangle BDF = area of triangle CDF.

Now draw a perpendicular from D onto BE and a perpendicular from A on to EF.

Looking at the angles AEF and BED, they are same. Angles AGE = angle DHE = 90 deg. Hence two triangles AEF and BED are similar. Further, side AE = DE. Hence two triangles are congruent. Hence, AG = DH.

Now see the two triangles ABF and DBF. Base BF is same and the altitudes AG and DH are same. Hence their areas are equal.

Hence, Area of ABC is divided into 3 equal parts BAF, BFD and CFD.

Area of AFB is 1/3 of area of ABC. Taking AC as the base of triangle ABC, we draw an altitude from B on to AC.

Then we will find that Since area of ABF is 1/3 area ABC, AF = 1/3 AC.

We join F and D.

In the two triangles BDF and CDF, wee see that BD = CD. The altitude from F on to BC will be same for both triangles. Hence Area of triangle BDF = area of triangle CDF.

Now draw a perpendicular from D onto BE and a perpendicular from A on to EF.

Looking at the angles AEF and BED, they are same. Angles AGE = angle DHE = 90 deg. Hence two triangles AEF and BED are similar. Further, side AE = DE. Hence two triangles are congruent. Hence, AG = DH.

Now see the two triangles ABF and DBF. Base BF is same and the altitudes AG and DH are same. Hence their areas are equal.

Hence, Area of ABC is divided into 3 equal parts BAF, BFD and CFD.

Area of AFB is 1/3 of area of ABC. Taking AC as the base of triangle ABC, we draw an altitude from B on to AC.

Then we will find that Since area of ABF is 1/3 area ABC, AF = 1/3 AC.