# NCERT Solutions For Class 9 Math Chapter – 7 Exercise – 7.1 NCERT Solutions For Class 9 Math Chapter – 7 Exercise – 7.1

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Q1.   In quadrilateral ABCD  , AC = AD and AB bisects A  Show that ABC   ̴ ∆ABD.  What can you say about BC and BD ?

Solution:- AB = AB [Common]

∟CAB = ∟DAB [ AB bisects ∟CAD]

∆ABC ̴ ∆ABD [SAS Rule]

Therefore BC = BD [CPCT]

Q2. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see figure) Prove that

• ABD BAC
• BD = AC
• ABD =

Solution:- • Consider triangles ABD and ABC.

We have AD = BC [Given]

AB = BA [Common]

And

∟DAB = ∟CBA [Given]

∆ABD  ∆BAC

BD = AC

∟ABD = ∟BAC [CPCT]

Q3. AD and BC are equal perpendicular to a line segment AB Show that CD bisects AB.

Solution:- AD and BC are perpendicular to AB.

AD // BC and CD is transversal.

OB = OA.

Hence , CD bisects AB.

Q4. l and M are two parallel lines intersected by another pair of parallel lines p and q Show thta ∆ABC  ∆CDA.

Solution:- ∟1 = ∟2 (I) [Alternate angles]

P // Q and AC is transversal

∟3 = ∟4  (ii) (Alternate angles)

AC = CA

∟1 = ∟2 and ∟3 = ∟4

∆ABC  ∆ ABC

AC = CA

∟1 = ∟2  and ∟3 = ∟4

∆ABC  ∆CDA

Q5. Line l is the bisector of an angle ∟A and B is any point on  l BP and BQ  are perpendicular from  to the arms of ∟A Show that:

• ∆APB AQB
• BP = BQ or B is equidistant from the arma of A

Solution:- Line l is the bisector of ∟QAP.

QAB = ∟PAB.

AB = PA

BQA = ∟BPA

• In triangles AQB and APB.

AB = BA and ∟BQA= ∟BPA

APB  ∆AQB.

• BP = BQ

Q6. In figures AC = AE , AB = AD and ∟BAD = ∟EAC Show that BC = DE.

Solution:- ∟BAD + ∟DAC = ∟DAC + ∟CAE   [ ∟DAC is added to both sides]

∟BAC = ∟DAE    (i)

In triangles ∆BAC and ∆DAE ,

AC = AE

∟BAC = ∟DAE

∆BAC   ∆DAC

BC = DE.

Q7. AB is a line segment and P is its mid- point D and E are points on the same sides of AB such that BAQ = ABE and EPA = DPB Show that

• ∆DAP EBP

Solution:- ∟EPA + ∟EPD = ∟EPD + ∟DPB

∟APD = ∟BPE

AP = BP

∟DAP = ∟EBP

• ∟APD = ∟BPE , AP = BP and ∟DAP = ∟EBP.

∆DAP ∆EBP.

Q8. 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

Show that:

• AMC BMD
• ∟DBC is a right angle.
• ∆DBC ACB
• CM = 1/2AB.

Solution:- • In ∆AMC and DMB ,

∟AMC = ∟BMD (VOA)

∆AMC  ∆BMD [ SAS rule]

• ∟BAC = ∟DBA [CPCT]

AB is the transversal

∟DBC is a right angle

• In ∆ABC and ∆DCB.

AC = DB [Since ∆AMC  ∆BMD ]

BC = CB [Common]

∟ACB = ∟DBC [Each 900]

∆DBC  ∆ACB [SAS rule]

• DC = AB [CPCT from part (iii]

2CM = AB

CM = 1/2AB.

CM  = 1/2AB.