Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB = ∠DBC. A D B C The figure below Q4 on page 14 — the single arcs mark ∠ABD and ∠DCA, the double arcs mark ∠DBC and ∠ACB.

Step 1 — add the given angles. At B the angle ∠ABC is made of ∠ABD and ∠DBC. At C the angle ∠DCB is made of ∠DCA and ∠ACB. ∠ABC = ∠ABD + ∠DBC ∠DCB = ∠DCA + ∠ACB Given ∠ABD = ∠DCA and ∠DBC = ∠ACB → ∠ABC = ∠DCB Step 2 — find the congruent triangles. Look at ∆ABC and ∆DCB, which share the side BC: ∠ABC = ∠DCB (just proved) BC = CB (common side) ∠ACB = ∠DBC (given) → ASA condition → ∆ABC ≅ ∆DCB Step 3 — list the equal parts. AB = DC and AC = DB (corresponding sides) ∠BAC = ∠CDB (corresponding angles) ∠ABC = ∠DCB, and of course BC = CB Step 4 — the crossing point. Let O be the point where AC and DB cross. In ∆OBC, ∠OBC = ∠OCB (these are the given equal angles ∠DBC and ∠ACB), so ∆OBC is isosceles and OB = OC, and therefore OA = OD (since AC = DB, and OA = AC − OC, OD = DB − OB) It also follows that ∆ABD ≅ ∆DCA by SSS: AB = DC, BD = CA and AD = DA. Why it happens: the whole figure is symmetric about the vertical line through the middle of BC. The two given equalities are exactly what forces that symmetry, and every pair listed above is a pair of parts swapped by it.

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

Book page 13–14 Updated on2026-09-19

Q1.

Identify whether the triangles below are congruent. What conditions did you use to establish their congruence? Express the congruence. (∆ABC with AB = 7 cm, BC = 5 cm, ∠B = 47°; ∆XYZ with XZ = 7 cm, ZY = 5 cm, ∠Z = 47°)

Answer

Yes, they are congruent, by the SAS condition.

In each triangle the marked angle lies between the two marked sides:

First triangle	Second triangle	Equal?
BA = 7 cm	ZX = 7 cm	Yes
∠B = 47° (between BA and BC)	∠Z = 47° (between ZX and ZY)	Yes
BC = 5 cm	ZY = 5 cm	Yes

So B corresponds to Z, A corresponds to X and C corresponds to Y:

∆ABC ≅ ∆XZY (SAS condition)

Why it happens: the vertex carrying the 47° angle must be matched with the vertex carrying the 47° angle, so B goes with Z. After that, the 7 cm arm goes with the 7 cm arm (A with X) and the 5 cm arm with the 5 cm arm (C with Y).

Tip: ∆ABC ≅ ∆XYZ would be wrong here. It would match ∠B with ∠Y, but the 47° angle in the second triangle is at Z, not at Y.

Q2.

Given that CD and AB are parallel, and AB = CD, what are the other equal parts in this figure? (Hint: When the lines are parallel, the alternate angles are equal. Are the two resulting triangles congruent? If so, express the congruence.)

The figure above Q3 on page 14 — DC and AB are the marked parallel sides, and DB and AC cross at O.

Answer

In the figure the segments DB and CA cross at O, making ∆ODC and ∆OBA.

Step 1 — the equal angles. DC and AB are parallel, so:

∠CDB = ∠ABD (alternate angles, transversal DB)
∠DCA = ∠BAC (alternate angles, transversal CA)

Step 2 — the congruence. Together with the given equal sides:

∠ODC = ∠OBA
DC = BA (given, the included side)
∠DCO = ∠BAO
→ ASA condition → ∆ODC ≅ ∆OBA

Step 3 — the other equal parts. From the congruence:

OD = OB
OC = OA
∠DOC = ∠BOA (also vertically opposite angles)

Why it happens: parallel lines hand us the two angles free of charge, and the given equal sides sit exactly between them. Once the triangles are congruent, the remaining sides must match too — which says that O is the midpoint of both DB and CA.

Q3.

Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that ∠BAC = ∠BDC. Are the two triangles congruent?

Answer

Yes, the triangles are congruent, and that is what makes the two angles equal.

The segment BC is shared by ∆ABC (above it) and ∆DBC (below it).

∠ABC = ∠DBC (given)
BC = BC (common side, included between the two angles)
∠ACB = ∠DCB (given)
→ ASA condition → ∆ABC ≅ ∆DBC

The correspondence is A ↔ D, B ↔ B, C ↔ C. Hence the third pair of angles must also match:

∠BAC = ∠BDC

Why it happens: the third angle of a triangle is decided by the other two, since all three add to 180°. Both triangles have the same two angles at B and at C, so their third angles must be equal as well. Congruence says the same thing in one step: corresponding angles of congruent triangles are equal.

Tip: because the triangles are congruent we also get AB = DB and AC = DC. So ABDC is a kite, with BC as its line of symmetry.

Q4.

Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB = ∠DBC.

The figure below Q4 on page 14 — the single arcs mark ∠ABD and ∠DCA, the double arcs mark ∠DBC and ∠ACB.

Answer

Step 1 — add the given angles. At B the angle ∠ABC is made of ∠ABD and ∠DBC. At C the angle ∠DCB is made of ∠DCA and ∠ACB.

∠ABC = ∠ABD + ∠DBC
∠DCB = ∠DCA + ∠ACB
Given ∠ABD = ∠DCA and ∠DBC = ∠ACB
→ ∠ABC = ∠DCB

Step 2 — find the congruent triangles. Look at ∆ABC and ∆DCB, which share the side BC:

∠ABC = ∠DCB (just proved)
BC = CB (common side)
∠ACB = ∠DBC (given)
→ ASA condition → ∆ABC ≅ ∆DCB

Step 3 — list the equal parts.

AB = DC and AC = DB (corresponding sides)
∠BAC = ∠CDB (corresponding angles)
∠ABC = ∠DCB, and of course BC = CB

Step 4 — the crossing point. Let O be the point where AC and DB cross. In ∆OBC, ∠OBC = ∠OCB (these are the given equal angles ∠DBC and ∠ACB), so ∆OBC is isosceles and

OB = OC, and therefore OA = OD
(since AC = DB, and OA = AC − OC, OD = DB − OB)

It also follows that ∆ABD ≅ ∆DCA by SSS: AB = DC, BD = CA and AD = DA.

Why it happens: the whole figure is symmetric about the vertical line through the middle of BC. The two given equalities are exactly what forces that symmetry, and every pair listed above is a pair of parts swapped by it.

Was this helpful?