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# In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

i) ar (BDE) =1/4 ar (ABC) ii) ar (BDE) = 1/2 ar (BAE)

iii) ar (ABC) = 2 ar (BEC) iv) ar (BFE) = ar (AFD)

v) ar (BFE) = 2 ar (FED) vi) ar (FED) = 1/8 ar (AFC)

[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]

**Solution:**

i) Let G and H be the mid-points of side AB and AC

Line segment GH is joining the mid-points and is parallel to the third side. Therefore, GH will be half of the length of BC (mid-point theorem).

∴ GH = 1/2 BC and GH || BD

∴ GH = BD = DC and GH || BD (D is the mid-point of BC)

Similarly,

- GD = HC = HA
- HD = AG = BG

Therefore, clearly, triangle ABC is divided into 4 equal equilateral triangles viz ΔBGD, ΔAGH, ΔDHC, and ΔGHD.

In other words,

ΔBGD = 1/4 ΔABC

Now consider ΔBDG and ΔBDE

BD = BD (Common base)

As both triangles are equilateral triangle, we can say BG = BE

DG = DE

Therefore,

ΔBDG ≅ ΔBDE [By SSS congruency]

The SSS rule states that: If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.

Thus, area (ΔBDG) = area (ΔBDE)

ar (ΔBDE) = 1/4 ar (ΔABC)

Hence proved.

ii) Join points A and D.

Area (ΔBDE) = Area (ΔAED) (Common base DE and DE || AB)

Area (ΔBDE) - Area (ΔFED) = Area (ΔAED) - Area (ΔFED)

Now Area (ΔBEF) = Area (ΔAFD) ... (1)

Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)

Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]

Area (ΔABD) = Area (ΔABE) ...(2)

AD is the median in ΔABC (D is the midpoint of BC)

ar (ΔABD) = 1/2 ar (ΔABC)

ar (ΔBDE) = 1/4 ar (ΔABC) [As proved earlier in (i)]

ar(ΔABD) = 2 ar (ΔBDE) ...(3)

From (2) and (3), we obtain,

2 ar (ΔBDE) = ar (ΔABE)

ar (BDE) = 1/2 ar (BAE)

iii) Join points C and E.

ar (ΔABE) = ar (ΔBEC) (Common base BE and BE || AC)

ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)

Using equation (1), we obtain,

ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)

ar (ΔABD) = ar (ΔBEC)

1/2 ar (ΔABC) = ar (ΔBEC)

ar (ΔABC) = 2 ar (ΔBEC)

iv) ΔBDE and ΔAED lie on the same base (DE) and between the parallels DE and AB.

ar (ΔBDE) = ar (ΔAED)

ar (ΔBDE) - ar (ΔFED) = ar (ΔAED) - ar (ΔFED) [Subtracting ar (ΔFED) on both the sides]

ar (ΔBFE) = ar (ΔAFD)

v) Let h be the height of vertex E, corresponding to the side BD in ΔBDE

Let H be the height of vertex A, corresponding to the side BC in ΔABC.

ar (ABC) = 1/2 × BC × H

ar (BDE) = 1/2 × 1/2(BC) × h

In (i), it was shown that ar (BDE) = 1/4 ar (ABC)

Therefore, H = 2h

Now,

ar (AFD) = 1/2 × FD × 2h

ar (FED) = 1/2 × (FD) × h

Hence, ar (AFD) = 2 × ar (FED)

In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD).

Therefore, ar (ΔBFE) = ar (ΔAFD) = 2 ar (ΔFED)

Hence proved.

vi) ar (ΔAFC) = ar (ΔAFD) + ar (ΔADC)

= 2 ar (ΔFED) + 1/2 ar (ΔABC) [using (v)]

= 2 ar (ΔFED) + 1/2 [4 × ar (ΔBDE)] [Using result of part (i)]

= 2 ar (ΔFED) + 2 ar (ΔBDE)

= 2 ar (ΔFED) + 2 ar (ΔAED)

[ΔBDE and ΔAED are on the same base and between same parallels]

= 2 ar (ΔFED) + 2 [ar (ΔAFD) + ar (ΔFED)]

= 2 ar (ΔFED) + 2 ar (ΔAFD) + 2 ar (ΔFED)

= 4 ar (ΔFED) + 4 ar (ΔFED) [using result of (v)]

⇒ ar (ΔAFC) = 8 ar (ΔFED)

⇒ ar (ΔFED) = 1/8 ar (ΔAFC)

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 9

**Video Solution:**

## In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that i) ar (BDE) =1/4 ar (ABC) ii) ar (BDE) = 1/2 ar (BAE) iii) ar (ABC) = 2 ar (BEC) iv) ar (BFE) = ar (AFD) v) ar (BFE) = 2 ar (FED) vi) ar (FED) = 1/8 ar (AFC). [Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.4 Question 5

**Summary:**

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC and AE intersects BC at F, then we can show that ar (BDE) = 1/4 ar (ABC), ar (BDE) = 1/2 ar (BAE), ar (ABC) = 2 ar (BEC),

ar (BFE) = ar (AFD) , ar (BFE) = 2 ar (FED), and ar (FED) = 1/8 ar (AFC).

**☛ Related Questions:**

- Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
- In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
- In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
- In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).[Hint: Join AC.]

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