In the figure below, are ∆DFE and ∆GED congruent to each other? It is given that DF = DG and FE = GE. D F G E The figure below Q4 on page 9 — DF = DG and FE = GE.

The two triangles in the figure are congruent — but ∆DFE ≅ ∆GED is not the correct way to write it . First, the congruence itself. Look at ∆DFE and ∆DGE: DF = DG (given) FE = GE (given) DE = DE (common side) → SSS condition → ∆DFE ≅ ∆DGE Now test the order given in the question. Writing ∆DFE ≅ ∆GED pairs the vertices like this: ∆DFE ∆GED Sides it matches Are they equal? D G DF with GE Not given F E FE with ED Not given E D ED with DG Not given None of these three pairs is known to be equal, so that correspondence is wrong. Why it happens: the two triangles really do fit over each other, but only when D stays on D and E stays on E, with F landing on G. Fold the figure along DE and the left half falls on the right half. So the honest statement is ∆DFE ≅ ∆DGE . Tip: ∆FDE ≅ ∆GDE and ∆EFD ≅ ∆EGD say the same thing. Always read the letters in pairs before you accept a congruence.

NCERT Solutions for Class 7th Maths Chapter 1 Figure it Out — Congruence of Triangles

Book page 8–9 Updated on2026-09-19

Q1.

Suppose ΔHEN is congruent to ΔBIG. List all the other correct ways of expressing this congruence.

Answer

The statement ∆HEN ≅ ∆BIG fixes three pairs of corresponding vertices:

H ↔ B, E ↔ I, N ↔ G

Any way of writing the same three pairs is correct. The three pairs can be listed in 6 orders, so besides the given one there are 5 other correct ways:

Way	Congruence	Pairs it records
Given	∆HEN ≅ ∆BIG	H–B, E–I, N–G
1	∆HNE ≅ ∆BGI	H–B, N–G, E–I
2	∆EHN ≅ ∆IBG	E–I, H–B, N–G
3	∆ENH ≅ ∆IGB	E–I, N–G, H–B
4	∆NHE ≅ ∆GBI	N–G, H–B, E–I
5	∆NEH ≅ ∆GIB	N–G, E–I, H–B

Why it happens: the congruence sign does not care which vertex is written first. It only demands that the first letter on the left goes with the first letter on the right, and so on. As long as the three pairs stay together, the statement means the same thing.

Tip: each of these can also be turned around, for example ∆BIG ≅ ∆HEN. Writing ∆HEN ≅ ∆BGI would be wrong — it pairs E with G and N with I.

Q2.

Determine whether the triangles are congruent. If yes, express the congruence. (∆RED with RE = 3.5 cm, ED = 5 cm, RD = 6 cm; and ∆JMA with JA = 3.5 cm, AM = 5 cm, JM = 6 cm)

Answer

Yes, they are congruent, by the SSS condition.

Match the equal sides:

Length	Side of the first triangle	Side of the second triangle
3.5 cm	RE	JA
5 cm	ED	AM
6 cm	RD	JM

Reading the matching down the columns gives R ↔ J, E ↔ A, D ↔ M. So

∆RED ≅ ∆JAM (SSS condition)

Why it happens: the three sidelengths of the two triangles are the same set — 3.5 cm, 5 cm, 6 cm — so SSS applies and the triangles are congruent. To write the congruence you must then find which vertex sits between which two sides. R lies between the 3.5 cm and 6 cm sides; in the second triangle that vertex is J.

Tip: writing ∆RED ≅ ∆JMA would be wrong. It would pair ED (5 cm) with MA — correct — but also RE (3.5 cm) with JM (6 cm), which is false.

Q3.

In the figure below, AB = AD, CB = CD. Can you identify any pair of congruent triangles? If yes, explain why they are congruent. Does AC divide ∠BAD and ∠BCD into two equal parts? Give reasons.

The figure below Q3 on page 9 — AB = AD and CB = CD, with AC drawn dashed.

Answer

Join AC. It splits the figure into ∆ABC and ∆ADC, and these two are congruent.

AB = AD (given)
CB = CD (given)
AC = AC (common side)
→ SSS condition → ∆ABC ≅ ∆ADC

Does AC divide the two angles equally? Yes.

Because the triangles are congruent, their corresponding angles are equal:

∠BAC = ∠DAC → AC cuts ∠BAD into two equal parts
∠BCA = ∠DCA → AC cuts ∠BCD into two equal parts

Why it happens: ∠BAC sits in ∆ABC at the vertex A, between the sides AB and AC. Its partner in ∆ADC sits at A between AD and AC. Since the triangles fit exactly over each other, these two angles must be equal — they are corresponding parts of congruent triangles. The same argument at C gives the second pair.

Did you know? A four-sided figure with two pairs of equal neighbouring sides like this is a kite. The diagonal AC is its line of symmetry — fold along AC and B lands on D.

Q4.

In the figure below, are ∆DFE and ∆GED congruent to each other? It is given that DF = DG and FE = GE.

The figure below Q4 on page 9 — DF = DG and FE = GE.

Answer

The two triangles in the figure are congruent — but ∆DFE ≅ ∆GED is not the correct way to write it.

First, the congruence itself. Look at ∆DFE and ∆DGE:

DF = DG (given)
FE = GE (given)
DE = DE (common side)
→ SSS condition → ∆DFE ≅ ∆DGE

Now test the order given in the question. Writing ∆DFE ≅ ∆GED pairs the vertices like this:

∆DFE	∆GED	Sides it matches	Are they equal?
D	G	DF with GE	Not given
F	E	FE with ED	Not given
E	D	ED with DG	Not given

None of these three pairs is known to be equal, so that correspondence is wrong.

Why it happens: the two triangles really do fit over each other, but only when D stays on D and E stays on E, with F landing on G. Fold the figure along DE and the left half falls on the right half. So the honest statement is ∆DFE ≅ ∆DGE.

Tip: ∆FDE ≅ ∆GDE and ∆EFD ≅ ∆EGD say the same thing. Always read the letters in pairs before you accept a congruence.

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