# Ncert Solutions For Class 9 Maths Ex 7.1

## Ncert Solutions For Class 9 Maths Chapter 7 Ex 7.1

Question 1:

In quadrilateral PQRS, PR = PQ and PQ bisects P$\angle P$ (look at the fig.). Show that the PQR = PQS$\triangle PQR\ =\ \triangle PQS$. What can you say about QR and QS?

Solution:

In PQR$\triangle PQR$ and PQS$\triangle PQS$ , we have

PR = PS

RPQ=SPQ$\angle RPQ = \angle SPQ$ (PQ bisects P$\angle P$)

PQ = PQ (common)

PQR=PQS$\triangle PQR = \triangle PQS$ (By SAS congruence)

Hence Proved.

Therefore, QR = QS (CPCT)

Question 2:

ABCD is a quadrilateral in which AD = BC and DAB$\angle DAB$= CBA$\angle CBA$ (see Fig.). Prove that

(i) ABD=BAC$\triangle ABD = \triangle BAC$

(ii) BD = AC

(iii) ABD=BAC$\angle ABD = \angle BAC$

Solution:

In the given figure, ABCD is a quadrilateral in which AD = BC and DAB$\angle DAB$= CBA$\angle CBA$.

In DAB and BAC$\triangle DAB\ and\ \triangle BAC$, we have

DAB=CBA$\angle DAB = \angle CBA$ [Given]

AB = AB [Common]

ABD=BAC$\triangle ABD = \triangle BAC$ [By SAS congruence]

BD = AC [CPCT]

and ABD=BAC$\angle ABD = \angle BAC$ [CPCT]

Hence Proved.

Question 3:

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

Solution:

In AOD=BOC$\triangle AOD = \triangle BOC$, we have,

AOD=BOC$\angle AOD = \angle BOC$ [Vertically opposite angles)

CBO=DAO(Each=90o)$\angle CBO = \angle DAO (Each = 90^{o})$

AOD=BOC$\triangle AOD = \triangle BOC$ [By AAS congruence]

Also, AO = BO [CPCT]

Hence, CD bisects AB Proved.

Question 4:

L and m are two para;;e; lines intersected by another pair of parallel lines p and q (see fig.). Show that ABC+CDA$\triangle ABC + \triangle CDA$ .

Solution:

In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., AB || DC and BC || AD.

In ABCandCDA$\triangle ABC \, and\,  \triangle CDA$, we have,

BAC=DCA$\angle BAC = \angle DCA$ (Alternate angles)

BCA=DAC$\angle BCA = \angle DAC$ (Alternate angles)

AC = AC (Common)

ABC=CDA$\triangle ABC = \triangle CDA$ (By SAS congruence)

Hence Proved.

Question 5:

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of A$\angle A$ (See fig.). Show that:

(i)  APB and AQB$\triangle APB\ and\ \triangle AQB$

(ii) BP = BQ or B is equidistant from the arms of A$\angle A$

Solution:

In APB and AQB$\triangle APB\ and\ \triangle AQB$ , we have

PAB=AQB$\angle PAB = \angle AQB$ (l is the bisector of A$\angle A$)

APB=AQB (Each=90o)$\angle APB = \angle AQB\ (Each = 90^{o})$

AB = AB

APB=AQB$\triangle APB = \triangle AQB$ (By AAS congruence)

Also, BP = BQ (By CPCT)

i.e, B is equidistant from the arms of A$\angle A$

Hence proved.

Question 6:

In the fig, AC = AE, AB = AD and BAD=EAC$\triangle BAD = \triangle EAC$ . Show that BC = DE.

Solution:

BAD=EAC$\angle BAD = \triangle EAC$ (Given)

BAD+DAC=EAC+DAC(AddingDAC to both sides)$\angle BAD + \angle DAC = \angle EAC + \angle DAC (Adding \angle DAC\ to\ both\ sides)$ BAC=EAC$\angle BAC = \angle EAC$

Now, in ABCandADE$\triangle ABC\, and\, \triangle ADE$ , We have

AC = AE

BAC=DAE$\angle BAC = \angle DAE$

ABC=ADE$\triangle ABC= \triangle ADE$ (By SAS congruence)

BC = DE (CPCT)

Hence Proved.

Question 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$\angle BAD = \angle ABE\ and\ \angle EPA = \angle DPB$ (see fig.). Show that

(i) DAP=EBP$\triangle DAP = \triangle EBP$

Solution:

In DAP and EBP$\triangle DAP\ and\ \triangle EBP$, we have

AP = BP (P is the midpoint of the line segment AB)

BAD=ABE$\angle BAD = \angle ABE$ (Given)

EPB=DPA$\angle EPB = \angle DPA$ (EPA=DPBEPA+DPE=DPB+DPE)$(\angle EPA = \angle DPB \Rightarrow \angle EPA + \angle DPE = \angle DPB + \angle DPE)$ DPA=EPB$\triangle DPA = \triangle EPB$

Hence Proved.

Question 8:

In right triangle ABC, right angles 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 fig.). Show that :

(i) AMC=BMD$\triangle AMC = \triangle BMD$

(II) DBC$\angle DBC$ is a right angle.

(iii) DBC=ACB$\triangle DBC = \triangle ACB$

(iv) CM=12AB$CM= \frac{1}{2} AB$

Solution:

In BMBandDMC$\triangle BMB and \triangle DMC$ , we have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

DMB=AMC$\angle DMB = \angle AMC$ (Vertically opposite angles)

AMC=BMD$\triangle AMC = \triangle BMD$ (By SAS)

Hence Proved.

(ii) AC || BD (DBM and CAM$( \angle DBM\ and\ \angle CAM$ are .alternate angles)

DBC+ACB=180o$\Rightarrow \angle DBC + \angle ACB = 180^{o}$ (Sum of co-interior angles)

DBC=90o$\Rightarrow \angle DBC = 90^{o}$

Hence proved.

(iii) In DBC amd ACB$\triangle DBC\ amd\ \triangle ACB$, We have

DB = AC (CPCT)

BC = BC (Common)

DBC=ACB(Each=90o)$\angle DBC = \angle ACB (Each = 90^{o})$

DBC=ACB$\triangle DBC = \triangle ACB$ (By SAS)

Hence proved.

(iv) AB = CD

12AB=12CD$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$ (CPCT)

Hence, 12AB=CM$\frac{1}{2} AB = CM$ Proved.