Question

17.
DE E F are the mid points of ABC, Show that angleABC = anglrDEF
angleACB=angleDFE and angleBAC =angleEDF​

Answer 1

Answer:

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 cm 2