# Triangles : Exercise 7.1 (Mathematics NCERT Class 9th)

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

Now in $\Delta s$ ABC and ABD, we have
$\angle CAB = \angle BAD$ [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) $\angle ABD = \angle BAC$

In $\Delta s$ ABD and BAC, we have
$\angle DAB = \angle CBA$ [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. Sol.

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 $\Delta ABC \cong \Delta CDA$. Sol.

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
Now, in $\Delta s$ ABC and CDA, we have
AB = CD [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 $\angle A$.

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. Sol.

In $\Delta s$ ABC and ADE, we have
$\angle BAC = \angle DAE$
[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) $\Delta \,DAP \cong \,\Delta EBP$

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) $CM = {1 \over 2}AB$

(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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