Question

ABC is an isosceles ∆, where AB =AC. E is any point on side AB. A line from E intersects BC produced at D. ED intersects AC at F.
Prove that:-
1) Angle AEF > Angle AFE
2) AF>AE

Answer 1

Given: In △ABC, D, E and F are midpoints of sides BC, CA and AB respectively.

Solution: Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and half of it.

Hence, DF=AC2

⇒DFAC=12−−−−−−(i)

Similarly, EFBC=12−−−−−−(ii)

and DEAB=12−−−−−−(iii)

From equation (i),(ii) and (iii), we have

DFAC=EFBC=DEAB=12

But if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

Hence, △ABC∼△DEF(By SSS similarity theorem)

⇒ar(△DEF)ar(△ABC)=EF2BC2

⇒ar(△DEF)ar(△ABC)=(12)2 (By using equation (ii))

⇒ar(△DEF)ar(△ABC)=14

⇒ar(△DEF)=ar(△ABC)4

⇒ar(△DEF)=204

⇒ar(△DEF)=5 cm2

Thanks for the request