- In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD. What can you say about BC and BD?Solution:
In ∆ABC and ∆ABD, we have AC = AD
∠CAB = ∠DAB [line AB intersects ∠CAB]
AB is common line segment
⸫ ∆ABC = ∆ABD, [By SAS congruence]
⸫ BC = AB ©
- ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) Δ ABD ≅ Δ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
In the given figure, ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA
In Δ ABD = Δ BAC, we have,
AD = BC (given)
∠DAB = ∠CBA (given)
AB = AB (common line segment)
⸫ Δ ABD ≅ Δ BAC (By SAS congruence)
⸫ BD = AC (Corresponding parts of congruent triangles are congruent)
and ∠ ABD = ∠ BAC (Corresponding parts of congruent triangles are congruent)
- AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
In ΔAOD and ΔBOC, we have,
∠AOD = ∠BOC [vertically opposite angles]
∠CBO = ∠DAO [given]
and AD = BC [given]
⸫ΔAOD ≅ ΔBOC [By AAS congruence]
Also, AO = BO (Corresponding parts of congruent triangles are congruent)
Hence CD bisects AB
- l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that Δ ABC ≅ Δ CDA.
Given l || m and p||q
To prove ΔABC ≅ ΔCDA
In ΔABC and ΔCDA
∠BCA = ∠DAC (alternate interior angles)
AC = CA ( common)
∠BAC = ∠DCA (alternate interior angles)
⸫ ΔABC ≅ ΔCDA (ASA congruence)
- line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Fig. 7.20). Show that:
(i) Δ APB ≅ Δ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.
I is the bisector of an ∠ A
BP and BQ are perpendiculars.
(i) In Δ APB and Δ AQB
∠P = ∠Q (right angles)
∠BAQ = ∠BAP (l is bisector)
AB = AB (common)
Therefore, Δ APB ≅ Δ AQB by AAS congruence condition.
(ii) BP = BQ by Corresponding parts of congruent triangles are congruent, B is equidistant from the arms of ∠A
- In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.
Given AC = AE , AB = AD and ∠BAD = ∠EAC
We have to prove, BC = DE
Let, ∠BAD = ∠EAC (Adding ∠DAC both sides)
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒∠BAC = ∠EAD
In, ΔABC and ΔADE,
AC = AE (given)
∠BAC = ∠EAD
AB = AD (Given)
Therefore, Δ ABC ≅ Δ ADE by SAS congruence condition
BC = DE by SAS congruence condition.
- AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Fig. 7.22). Show that
(i) Δ DAP ≅ Δ EBP
(ii) AD = BE
Given P is the midpoint of AB
∠BAD = ∠ABE and ∠EPA = ∠DPB
∠EPA = ∠DPB (adding ∠DPE both sides)
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒∠DPA = ∠EPB
In, Δ DAP ≅ Δ EBP,
∠DPA = EPB
AP = BP (P is the midpoint of AB)
∠BAD = ∠ABE (given)
Therefore, Δ DAP ≅ Δ EBP by ASA congruence condition
⇒AD = BE (Corresponding parts of congruent triangles are congruent)
- In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:
(i) Δ AMC ≅ Δ BMD
(ii) ∠ DBC is a right angle.
(iii) Δ DBC ≅ Δ ACB
(iv) CM = 1/2 AB
(i) In ΔBMD and ΔAMC,
AM = BM (M is the midpoint)
∠CMA = ∠DMB (vertically opposite angles)
CM = DM (given)
⇒ ΔAMC ≅ ΔBMD by SAS congruence condition.
(ii) ∠ACM = ∠BDM (by Corresponding parts of congruent triangles are congruent)
Therefore, AC||BD as alternate interior angles are equal.
Now, ∠ACB + ∠DBC = 180˚
90˚ + ∠B = 180˚
∠DBC = 90˚
(iii) in ΔDBC and ΔACB
BC = CB (common)
∠ACB = ∠DBC (right angles)
DB = AC (byCorresponding parts of congruent triangles are congruent)
⸫ Δ DBC ≅ Δ ACB by SAS congruence condition.
(iv) DC = AB (ΔDBC ≅ ΔACB)
DM = CM = AM = BM (M is midpoint)
⇒ DM + CM = AM + BM
⇒ DM + CM = AB
⇒ CM = 1/2 AB
Next exercise – Triangles – Exercise 7.2 – Class IX