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Triangles : Exercise 7.1 (Mathematics NCERT Class 9th)


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Q.1     In quadrilateral  ACBD, AC = AD and AB bisects \angle A (see figure). Show that \Delta \,ABC \cong \,\Delta ABDWhat can you say about BC and BD?

1Sol.

Now in \Delta s ABC and ABD, we have  
AC = AD [Given]
\angle CAB = \angle BAD [Since AB bisects \angle A]
and AB = AB [Common]
Therefore by SAS congruence criterion, we have
\Delta ABC \cong \Delta ABD
 \Rightarrow BC = BD [Since corresponding parts of congruent triangles are equal]


Q.2      ABCD is a quadrilateral in which AD = BC and \angle DAB = \angle CBA (see figure). Prove that 
           
(i) \Delta ABD \cong \Delta BAC
            (ii) BD = AC
            (iii) \angle ABD = \angle BAC

3Sol.

In \Delta s ABD and BAC, we have
AD = BC [Given]
\angle DAB = \angle CBA [Given]
AB = AB [Common]
Therefore by SAS criterion of congruence, we have
\Delta ABD \cong \,\Delta BAC,, which proves (i)
 \Rightarrow BD = AC
and , \angle ABD = \angle BAC, which proves (ii) and (iii)
[Since corresponding parts of congruent triangles are equal]


Q.3      AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

2Sol.

Since AB and CD intersect at O. Therefore ,
\angle AOD = \angle BOC .... (1) [Vertically opp. angles]
In \Delta s AOD and BOC, we have
\angle AOD = \angle BOC [From (1)]
\angle DAO = \angle OBC [Each = 90º]
and , AD = BC [Given]
Therefore by AAS congruence criterion, we have
\Delta AOD \cong \Delta BOC
 \Rightarrow OA = OB
[Since corresponding parts of congruent triangles are equal]
i.e., O is the mid- point of  AB.
Hence, CD bisects AB.


Q.4      l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that \Delta ABC \cong \Delta CDA.

5Sol.

Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore , AD || BC and AB || CD
 \Rightarrow ABCD is a  parallelogram.

i.e. ,  AB = CD
and BC = AD
Now, in \Delta s ABC and CDA, we have
AB = CD [Proved above]
BC = AD [Proved above]
and AC = AC [Common]
Therefore By SSS criterion of congruence.
\Delta ABC \cong \Delta CDA


Q.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 \angle A (see figure). Show that : 
          
(i) \Delta APB \cong \Delta AQB
           (ii) BP = BQ or B is equidistant from the arms of \angle A.

4Sol.

In \Delta s APB and AQB we have
\angle APB = \angle AQB [Since Each = 90º]
\angle PAB = \angle QAB [Since AB bisects \angle PAQ]
AB = AB [Common]
By AAS congruence criterion, we have
\Delta \,APB \cong \,\Delta AQB, which proves (i)
 \Rightarrow BP = PQ
[Since Corresponding parts of congruent triangles are equal]
i.e., B is equidistant from the arms of \angle A, which proves (ii).


Q.6       In figure AC = AE, AB = AD and \angle BAD = \angle EAC. Show that BC = DE.

6Sol.

In \Delta s ABC and ADE, we have
AB = AD [Given]
\angle BAC = \angle DAE
[Since \angle BAD = \angle EAC  \Rightarrow \angle BAD + \angle DAC = \angle EAC + \angle DAC  \Rightarrow
 \Rightarrow \angle BAD = \angle DAE]
and , AC = AE   [Given ]
Therefore By SAS criterion of congruence, we have
\Delta \,ABC \cong \,\Delta ADE
 \Rightarrow BC = DE
[Since Corresponding parts of congruent triangles are equal]


Q.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 \angle BAD = \angle ABE and \angle EPA = \angle DPB (see figure). Show that - 
            (i) \Delta \,DAP \cong \,\Delta EBP

            (ii) AD = BE.

7Sol.

We have , \angle EPA = \angle DPB
 \Rightarrow \angle EPA + \angle DPE = \angle DPB + \angle DPE
 \Rightarrow \angle DPA = \angle EPB
Now , in \Delta s EBP and DAP, we have
\angle EPB = \angle DPA [From (1)]
BP = AP                                                [Given]
and , \angle EBP = \angle DAP [Given]
So, by ASA criterion of congruence, we have
\Delta \,EBP\, \cong \,\Delta DAP
 \Rightarrow BE = AD i.e., AD = BE
[Since Corresponding parts of congruent triangles are equal. ]


Q.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) \Delta \,AMC\, \cong \,\Delta \,BMD
            (ii) \angle DBC is a right angle
            (iii) \Delta \,DBC\, \cong \,\Delta \,ACB
            (iv) CM = {1 \over 2}AB

8Sol.

(i) In \Delta s AMC and BMD, we have
AM = BM [Since M is the mid- point of AB]
\angle AMC = \angle BMD [Vertically opp. \angle s]
and CM = MD [Given]
Therefore By SAS criterion of congruence , we have
\Delta \,AMC \cong \Delta BMD

(ii) Now, \Delta \,AMC \cong \Delta BMD
 \Rightarrow BD = CA\,and\,\,\angle BDM = \angle ACM ... (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 \angle BDM\,and\,\angle ACM are equal. Therefore , BD || CA.
 \Rightarrow \angle CBD\, + \,\angle BCA = 180º
[Since sum of consecutive interior angles are supplementary ]
 \Rightarrow \angle CBD\, + 90º = 180º [Since \angle BCA = 90º]
 \Rightarrow \angle CBD\, = 180º - 90º
 \Rightarrow \angle DBC = 90º

(iii) Now, in \Delta s DBC and ACB, we have
BD = CA [From (1)]
\angle DBC = \angle ACB [Since Each = 90º]
BC = BC [Common]
Therefore by SAS criterion of congruence, we have
\Delta \,DBC \cong \,\Delta ACB

(iv) CD = AB [Since corresponding parts of congruent triangles are equal]
 \Rightarrow {1 \over 2}CD = {1 \over 2}AB  \Rightarrow CM = {1 \over 2}AB



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