Class 9
Explaination needed
Q.1: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Q.2: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Q.3: 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. Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Q.4: 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. Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE
Q.5: 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 the figure). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2 AB
Answers
Step-by-step explanation:
Answers:-
Q-1:-
Given that
AD = BC
angle DAB = angle CBA
now in ∆ABD and ∆BAC
AB=AB (Common side)
AD = BC
angle DAB= angle CBA
ΔABD ≅ ΔBAC
by SAS Property
AC = BD
Congruent parts in the Congruent triangles
angle ABD = angle BAC
Q-2:-
Given that:
AD = BC
AD is perpendicular to AB and
BC is perpendicular to AB.
In ∆BOC and ∆AOD
Angle OBC = angle AOD
(Vertically opposite angles)
angle OBC = angle OAD(right angles)
BC=AD
∆OBC is congruent to ∆OAD
by AAS Property
=> OB = OA
Congruent parts in the Congruent triangles
=> O bisects AB----(1)
And OD =OC
=> O bisects CD-----(2)
From (1)&(2)
=> AB bisects CD
Hence , Proved.
Q-5:-
Join A and D
Given that
angle C = 90°
M is the mid point of AB
MD = MC
1)
In ∆AMC and ∆BMD
AM = BM
CM=DM
angle AMC = angle BMD
(Vertically opposite angles)
ΔAMC ≅ ΔBMD
By SAS property
2)
by the above we have
angle MDB = angle MCA
(alternative interior angles)
=> BD || AC
We have AC perpendicular to BC
=>BD is perpendicular to BC
angle DBC =90°
3)
In ∆DBC and ∆ACB
BD = AC
angle DBC = angle ACB =90°
BC=BC (common side)
ΔDBC ≅ ΔACB
By SAS Property
4)
DC = AB
(1/2)DC = (1/2)AB
=> CM = (1/2)AB
CM = AB/2