In the adjoining figure Angle ACD = 30° and angle ABD =60° , AB = AD
Find angle BAD , angle DAC ,angle BAC
Answers
Answer:
1. In quadrilateral ACBD, AC = AD and AB bisects ∠A )see figure). Show that ΔABC ≌ ΔABD. What can you say about BC and BD?
Ans. In quadrilateral ABCD we have
AC = AD
and AB being the bisector of ∠A.
Now, in ΔABC and ΔABD,
AC = AD
[Given]
AB = AB
[Common]
∠CAB = ∠DAB [∴ AB bisects ∠CAD]
∴ Using SAS criteria, we have
ΔABC ≌ ΔABD.
∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.
∴ 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.
Ans. (i) In quadrilateral ABCD, we have AD = BC and
∠DAB = ∠CBA.
In ΔABD and ΔBAC,
AD = BC
[Given]
AB = BA
[Common]
∠DAB = ∠CBA
[Given]
∴ Using SAS criteria, we have ΔABD ≌ ΔBAC
(ii) ∵ ΔABD ≌ ΔBAC
∴ Their corresponding parts are equal.
⇒ BD = AC
(ii) Since ΔABD ≌ ΔBAC
∴ Their corresponding parts are equal.
⇒ ∠ABD = ∠BAC.
3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Ans. We have ∠ABC = 90° and ∠BAD = 90°
Also AB and CD intersect at O.
∴ Vertically opposite angles are equal.
Now, in ΔOBC and ΔOAD, we have
∠ABC = ∠BAD
[each = 90°]
BC = AD
[Given]
∠BOC = ∠AOD
[vertically opposite angles]
∴ Using ASA criteria, we have
ΔOBC ≌ ΔOAD
⇒ OB = OA
[c.p.c.t]
i.e. O is the mid-point of AB
Thus, 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.
Ans. ∵l || m and AC is a transversal.
∴ ∠BAC = ∠DCA
[Alternate interior angles]
Also p || q and AC is a transversal,
∴ ∠BCA = ∠DAC
[Alternate interior angles]
Now, in ΔABC and ΔCDA,
∠BAC = ∠DCA
[Proved]
∠BCA = ∠DAC
[Proved]
CA = AC
[Common]
∴ Using ASA criteria, we have ΔABC ≌ ΔCDA
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) BO = BQ or B is equidistant from the arms of ∠A.
Ans. We have, l as the bisector of QAP.
∴ ∠QAB = ∠PAB
∴ ∠Q = ∠P
[each = 90°]
⇒ Third ∠ABQ = Third ∠ABP
(i) Now, in ΔAPB and ΔAQB, we have
AB = AB
[Common]
∠ABP = ∠ABQ
[Proved]
∠PAB = ∠QAB
[Proved]
∴ Using SAS criteria, we have
ΔAPB ≌ ΔAQB
(ii) Since ΔAPB ≌ ΔAQB
∴ Their corresponding angles are equal.
⇒ BP = BQ
i.e. [Perpendicular distance of B from AP]
= [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A.
6. In the figure, AC = AE, AB = AD and ∠BAD = ∠ESC. Show that BC = DE.
Ans. We have ∠BAD = ∠EAC
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ΔABC and ΔADE, we have
∠BAC = ∠DAE
[Proved]
AB = AD
[Given]
AC = AE
[Given]
∴ ΔABC ≌ ΔADE
[Using SAS criteria]
Since ΔABC ≌ ΔADE, therefore, their corresponding parts are equal.
⇒ BC = DE.
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
Ans. We have, P is the mid-point of AB.
AP = BP
∠EPA = ∠DPB
[Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ APD = ∠BPE
(i) Now, in ΔDAP ≌ ΔEBP, we have
AP = BP
[Proved]
∠PAD = ∠ PBE
[∴ It is given that ∠BAD = ∠ABE]
∠DPA = ∠EPB
[Proved]
∴ Using ASA criteria, we have
ΔDAP ≌ ΔEBP
(ii) Since,
∴ Their corresponding parts are equal.
⇒ AD = BE.
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
(ii) ∠DBC is a right angle.
(ii) ΔDBC ≌ ΔACB
(iv) CM = AB
Ans. ∵ M is the mid-point of AB.
∴ BM = AM
[Given]
(i) In ΔAMC and ΔBMD, we have
CM = DM
[Given]
AM = BM
[Proved]
∠AMC = ∠BMD
[Vertically opposite angles]
∴ ΔAMC ≌ ΔBMD
[SAS criteria]
(ii) ∴ΔAMC ≌ ΔBMD
∴ Their corresponding parts are equal.
⇒ ∠MAC = ∠MBD
But they form a pair of alternate interior angles.
∴ AC || DB
Now, BC is a transversal which intersecting parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180°
But ∠BCA = 90°
[∴ ΔABC is right angled at C]
∴ 90° + ∠DBC = 180°
or ∠DBC = 180° – 90° = 90°
Thus, ∠DBC = 90°
(iii) Again, ΔAMC ≌ ΔBMD
[Proved]
∴ AC = BD
[c.p.c.t]
Now, in ΔDBC and ΔACB, we have
∠DBC = ∠ACB
[Each = 90°]
BD = CA
[Proved]
BC = CB
[Common]
∴ Using SAS criteria, we have
ΔDBC ≌ ΔACB
(iv) ∴ ΔDBC ≌ ΔACB
∴ Their corresponding parts are equal.
⇒ DC = AB
But DM = CM
[Given]