Q.1Â Â Â In quadrilateral Â ACBD, AC = AD and AB bisects (see figure). Show that .Â What can you say about BC and BD?
Now in ABC and ABD, we have Â
AC = AD [Given]
[Since AB bisects ]
and AB = AB [Common]
Therefore by SAS congruence criterion, we have
BC = BD [Since corresponding parts of congruent triangles are equal]
Q.2Â Â Â Â Â ABCD is a quadrilateral in which AD = BC and (see figure). Prove thatÂ
Â Â Â Â Â Â Â Â Â Â Â (i)
Â Â Â Â Â Â Â Â Â Â Â (ii) BD = AC
Â Â Â Â Â Â Â Â Â Â Â (iii)
In ABD and BAC, we have
AD = BC [Given]
[Given]
AB = AB [Common]
Therefore by SAS criterion of congruence, we have
, which proves (i)
BD = AC
and , which proves (ii) and (iii)
[Since corresponding parts of congruent triangles are equal]
Q.3Â Â Â Â Â AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Since AB and CD intersect at O. Therefore ,
.... (1) [Vertically opp. angles]
In AOD and BOC, we have
[From (1)]
[Each = 90Âº]
and , AD = BC [Given]
Therefore by AAS congruence criterion, we have
OA = OB
[Since corresponding parts of congruent triangles are equal]
i.e., O is the mid- point of Â AB.
Hence, CD bisects AB.
Q.4Â Â Â Â Â l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that .
Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore , AD || BC and AB || CD
ABCD is a Â parallelogram.
i.e. ,Â AB = CD
and BC = AD
Now, in ABC and CDA, we have
AB = CD [Proved above]
BC = AD [Proved above]
and AC = AC [Common]
Therefore By SSS criterion of congruence.
Q.5Â Â Â Â 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 (see figure). Show that :Â
Â Â Â Â Â Â Â Â Â Â (i)
Â Â Â Â Â Â Â Â Â Â (ii) BP = BQ or B is equidistant from the arms of .
In APB and AQB we have
[Since Each = 90Âº]
[Since AB bisects ]
AB = AB [Common]
By AAS congruence criterion, we have
, which proves (i)
BP = PQ
[Since Corresponding parts of congruent triangles are equal]
i.e., B is equidistant from the arms of which proves (ii).
Q.6Â Â Â Â Â Â In figure AC = AE, AB = AD and Show that BC = DE.
In ABC and ADE, we have
AB = AD [Given]
[Since
]
and , AC = AEÂ Â [Given ]
Therefore By SAS criterion of congruence, we have
BC = DE
[Since Corresponding parts of congruent triangles are equal]
Q.7Â Â Â Â Â AB is a line segment and P is its mid- point. D and E are points on the same side of AB such that andÂ (see figure). Show that -Â
Â Â Â Â Â Â Â Â Â Â Â (i)
Â Â Â Â Â Â Â Â Â Â Â (ii) AD = BE.
We have ,
Now , in EBP and DAP, we have
[From (1)]
BP = AP Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]
and , [Given]
So, by ASA criterion of congruence, we have
BE = AD i.e., AD = BE
[Since Corresponding parts of congruent triangles are equal. ]
Q.8Â Â Â Â 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 :
Â Â Â Â Â Â Â Â Â Â Â (i)
Â Â Â Â Â Â Â Â Â Â Â (ii) is a right angle
Â Â Â Â Â Â Â Â Â Â Â (iii)
Â Â Â Â Â Â Â Â Â Â Â (iv)
(i) In AMC and BMD, we have
AM = BM [Since M is the mid- point of AB]
[Vertically opp. ]
and CM = MD [Given]
Therefore By SAS criterion of congruence , we have
(ii) Now,
... (1)
[Since corresponding parts of congruent triangles are equal]
Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate angles are equal. Therefore , BD || CA.
= 180Âº
[Since sum of consecutive interior angles are supplementary ]
90Âº = 180Âº [Since = 90Âº]
= 180Âº - 90Âº
90Âº
(iii) Now, in DBC and ACB, we have
BD = CA [From (1)]
[Since Each = 90Âº]
BC = BC [Common]
Therefore by SAS criterion of congruence, we have
(iv) CD = AB [Since corresponding parts of congruent triangles are equal]
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