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In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE . Prove that BD/CD = BF/CE

In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE . Prove that BD/CD = BF/CE

Solution:

Given, line segment DF intersects the side AC of a triangle ABC at E.

E is the midpoint of CA

Also, ∠AEF = ∠AFE

We have to prove that BD/CD = BF/CE

Since E is the midpoint of CA, then

CE = AE --------------- (1)

Given, ∠AEF = ∠AFE ---------------- (2)

So, AE = AF ------------------- (3)

From (1) and (3),

CE = AE = AF --------------- (4)

Now, draw CG parallel to DF,

In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE . Prove that BD/CD = BF/CE

From the figure,

The corresponding angles are equal. i.e., ∠AEF = ∠ACG ------------ (5)

Also, ∠AFE = ∠AGC ---------------- (6)

From (2), (5) and (6),

∠AFE = ∠AEF = ∠ACG = ∠AGC

We know that sides opposite to equal angles are equal.

So, AC = AG ---------------- (7)

We know, AC = CE + AE

AG = AF + GF

So, (7) becomes CE + AE = AF + GF

From (4) and (7),

AE = CE = AF = GF

CE = AF = GF ----------------- (8)

Basic Proportionality Theorem(BPT) states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

By BPT,

CB/DC = BG/GF

Adding 1 on both sides,

CB/DC + 1 = BG/GF + 1

CB+DC / DC = BG+GF / GF

From the figure,

CB + CD = BD

BG + GF = BF

So, BD/DC = BF/GF

Therefore, it is proven that BD/DC = BF/GF

✦ Try This: The side BC of a triangle ABC is bisected at D; O is any point in AD,BO and CO produced to meet AC and AB in E and F respectively and AD is produced to X so that D is the mid-point of OX. Prove that AO:AX = AF:AB and show that FE || BC.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6

NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 16

In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE . Prove that BD/CD = BF/CE

Summary:

In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE . It is proved that BD/CD = BF/CE

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