TEXTBOOK QUESTIONS SOLVED
1. In quadrilateral ABCD, AC = AD and AB bisects ∠A (see figure).
Show that ΔABC ≅ ΔABD.
What can you say about BC and BD?
Consider triangles ABC and ABD,
We have AC = AD [Given]
AB = AB [Common]
and ∠CAB = ∠DAB [ ∴ AB bisects ∠CAD]
ΔABC ≅ ΔABD [SAS rule]
Therefore, BC = BD.
2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
(i) Consider triangles ABD and ABC,
We have AD = BC [Given]
AB = BA [ Commo ]
and ∠DAB = ∠CBA [ Given ]
∴ ΔABD ≅ ΔBAC [ SAS rule]
(ii) BD = AC. [ CPCT ]
(iii) LABD = LBAC, [ CPCT ]
3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Sol. AD and. BC are perpendiculars to AB.
and CD is transversal.
∴ ∠BCD = ∠ADC, i.e.. ∠BCO = ∠ADO [ Alternate agnles ]
Consider triangles BCO and ADO,
We have BC = AD [ Given ]
∠ OBC = ∠OAD [90° each ]
and ∠BCO = ∠ADO [ Proved above ]
∴ ΔOBC ≅ ΔOAD [ CPCT ]
Therefore, OB = OA
Hence, CD bisects AB.
4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC ≅ ΔCDA.
and AC is transversal.
So, ∠ 1 = ∠ 2 ….. (i) [Alternate angles]
Again, and AC is transversal.
So, ∠ 3 = ∠ 4 …(ii) [Alternate angles]
Now, in ΔADC and ΔABC,
AC = CA [ Common ]
∠ 1 = ∠2 and ∠ 3 = ∠ 4 [ From (i) and (ii) ]
∴ ΔABC ≅ Δ CDA [ASA rule ]
5. 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 figure). Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
is the bisector of ΔQAP.
∴ ∠QAB = ∠PAB …(i)
AB = BA …(ii) [Common]
and ∠BQA = ∠BPA. …(iii) [90° each]
(i) In triangles AQB and APB,
∠QAB = ∠PAB ; AB = BA and ∠BQA = ∠BPA. [From (i), (ii), (iii)]
∴ ∠APB ≅ ∠AQB. [AAS rule]
(ii) Therefore, BP = BQ [CPCT]
i.e., B is equidistant from the arms of ∠A.
6. In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
As ∠BAD = ∠CAE [Given]
⇒ ∠BAD + ∠DAC
= ∠DAC + ∠CAE
[∠ DAC is added to both sides]
⇒ ∠BAC = ∠DAE
Consider triangles BAC and DAE,
We have AB = AD [ Given ]
AC =AE [ Given ]
and ∠BAC = ∠DAE [ SAS rule ]
⇒ BC = DE. [ CPCT ]
7. 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 figure). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE.
Since ∠EPA = ∠DPB [Given]
⇒ ∠EPA + ∠EPD = ∠EPD + ∠DPB [Adding ∠EPD to both sides]
⇒ ∠APD = ∠BPE …(i)
Also, AP = BP . ..(ii) [Given]
and ∠DAP = ∠EBP …(iii) [Given]
(i) Consider triangles DAP and EBP,
∠APD = ∠BPE, AP = BP and ∠DAP = ∠EBP.
[From (i), (ii), (iii)]
ΔDAP ≅ ΔEBP. [SAS rule]
(ii) AD = BE [ CPCT ]
8. 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 figure). Show that:
(i) ΔAMC ≅ ΔBMD
– Consider triangles AMC and DMB,
We have AM = BM [Given]
CM = DM [Given]
and ∠AMC = ∠BMD [Vertically opposite angles]
∴ Δ AMC ≅ Δ BMD [SAS rule]
(ii) ∠ DBC is a right angle.
Sol. – As ∠BAC = ∠DBA [CPCT, from part (i)]
and AB is the transversal.
We have …(ii) [Given]
So, [From (1) and (ii)]
i.e., ∠ DBC is a right angle, i.e., ∠ DBC = 90°.
(iii) Δ DBC ≅ ΔACB
Sol. – Consider triangles ABC and DCB.
We have AC = DB [Since ΔAMC ≅ Δ BMD; result (i)]
BC = CB [Common]
and ∠ACB = ∠DBC [Each 90° ]
∴ ΔDBC ≅ ΔACB [SAS rule]
Sol. – As DC = AB [CPCT from part (iii)]
⇒ 2CM = AB [ ∴ M is mid-point of DC]