**Chapter
7 Triangles**

**Ex 7.1 Class 9 Maths Question
1.**

**In quadrilateral ACBD, AC = AD and AB bisects**

**∠**

**A (see figure). Show that ∆ABC**

**≅**

**∆ABD. What can you say about BC and BD?**

Solution:

In quadrilateral ACBD, we have AC = AD and AB being the bisector of

Solution:

In quadrilateral ACBD, we have AC = AD and AB being the bisector of

**∠**

**A.**

Now, In ∆ABC and ∆ABD,

AC = AD (Given)

Now, In ∆ABC and ∆ABD,

AC = AD (Given)

**∠**

**CAB =**

**∠**

**DAB ( AB bisects**

**∠**

**CAB)**

and AB = AB (Common)

and AB = AB (Common)

**∴**

**∆ ABC**

**≅**

**∆ABD (By SAS congruence axiom)**

**∴**

**BC = BD (By CPCT)**

**Ex 7.1 Class 9 Maths Question 2.**

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

(i) ∆ABD

(i) ∆ABD

**≅**

**∆BAC**

(ii) BD = AC

(iii)

(ii) BD = AC

(iii)

**∠**

**ABD =**

**∠**

**BAC**

Solution:

In quadrilateral ACBD, we have AD = BC and

Solution:

In quadrilateral ACBD, we have AD = BC and

**∠**

**DAB =**

**∠**

**CBA**

**(i) In
∆ ABC and ∆ BAC,
AD = BC (Given)
**

**∠**

**DAB =**

**∠**

**CBA (Given)**

AB = AB (Common)

AB = AB (Common)

**∴**

**∆ ABD**

**≅**

**∆BAC (By SAS congruence)**

**(ii)
Since ∆ABD ****≅**** ∆BAC
**

**⇒**

**BD = AC [By C.P.C.T.]**

**(iii)
Since ∆ABD ****≅**** ∆BAC
**

**⇒**

**∠**

**ABD =**

**∠**

**BAC [By C.P.C.T.]**

**Ex 7.1 Class 9 Maths Question
3.**

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

Solution:

In ∆BOC and ∆AOD, we have

Solution:

In ∆BOC and ∆AOD, we have

**∠**

**BOC =**

**∠**

**AOD**

BC = AD [Given]

BC = AD [Given]

**∠**

**BOC =**

**∠**

**AOD [Vertically opposite angles]**

**∴**

**∆OBC**

**≅**

**∆OAD [By AAS congruency]**

**⇒**

**OB = OA [By C.P.C.T.]**

i.e., O is the mid-point of AB.

Thus, CD bisects AB.

i.e., O is the mid-point of AB.

Thus, CD bisects AB.

**Ex 7.1 Class 9 Maths Question
4.**

**l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.**

Solution:

Solution:

**∵**

**p || q and AC is a transversal,**

**∴**

**∠**

**BAC =**

**∠**

**DCA …(1) [Alternate interior angles]**

Also l || m and AC is a transversal,

Also l || m and AC is a transversal,

**∴**

**∠**

**BCA =**

**∠**

**DAC …(2)**

[Alternate interior angles]

Now, in ∆ABC and ∆CDA, we have

[Alternate interior angles]

Now, in ∆ABC and ∆CDA, we have

**∠**

**BAC =**

**∠**

**DCA [From (1)]**

CA = AC [Common]

CA = AC [Common]

**∠**

**BCA =**

**∠**

**DAC [From (2)]**

**∴**

**∆ABC**

**≅**

**∆CDA [By ASA congruency]**

**Ex 7.1 Class 9 Maths Question
5.**

**Line l is the bisector of an**

**∠**

**A and**

**∠**

**B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that**

(i) ∆APB

(i) ∆APB

**≅**

**∆AQB**

(ii) BP = BQ or B is equidistant from the arms ot

(ii) BP = BQ or B is equidistant from the arms ot

**∠**

**A.**

Solution:

We have, l is the bisector of

Solution:

We have, l is the bisector of

**∠**

**QAP.**

**∴**

**∠**

**QAB =**

**∠**

**PAB**

**∠**

**Q =**

**∠**

**P [Each 90°]**

**∠**

**ABQ =**

**∠**

**ABP**

[By angle sum property of A]

Now, in ∆APB and ∆AQB, we have

[By angle sum property of A]

Now, in ∆APB and ∆AQB, we have

**∠**

**ABP =**

**∠**

**ABQ [Proved above]**

AB = BA [Common]

AB = BA [Common]

**∠**

**PAB =**

**∠**

**QAB [Given]**

**∴**

**∆APB**

**≅**

**∆AQB [By ASA congruency]**

Since ∆APB

Since ∆APB

**≅**

**∆AQB**

**⇒**

**BP = BQ [By C.P.C.T.]**

i. e., [Perpendicular distance of B from AP]

= [Perpendicular distance of B from AQ]

Thus, the point B is equidistant from the arms of

i. e., [Perpendicular distance of B from AP]

= [Perpendicular distance of B from AQ]

Thus, the point B is equidistant from the arms of

**∠**

**A.**

**Ex 7.1 Class 9 Maths Question
6.**

**In figure, AC = AE, AB = AD and**

**∠**

**BAD =**

**∠**

**EAC. Show that BC = DE.**

Solution:

We have,

Solution:

We have,

**∠**

**BAD =**

**∠**

**EAC**

Adding

Adding

**∠**

**DAC on both sides, we have**

**∠**

**BAD +**

**∠**

**DAC =**

**∠**

**EAC +**

**∠**

**DAC**

**⇒**

**∠**

**BAC =**

**∠**

**DAE**

Now, in ∆ABC and ∆ADE. we have

Now, in ∆ABC and ∆ADE. we have

**∠**

**BAC =**

**∠**

**DAE [Proved above]**

AB = AD [Given]

AC = AE [Given]

AB = AD [Given]

AC = AE [Given]

**∴**

**∆ABC**

**≅**

**∆ADE [By SAS congruency]**

**⇒**

**BC = DE [By C.P.C.T.]**

**Ex 7.1 Class 9 Maths Question
7.**

**AS 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. (see figure). Show that**

(i) ∆DAP

(i) ∆DAP

**≅**

**∆EBP**

(ii) AD = BE

(ii) AD = BE

Solution:

We have, P is the mid-point of AB.

Solution:

We have, P is the mid-point of AB.

**∴**

**AP = BP**

**∠**

**EPA =**

**∠**

**DPB [Given]**

Adding

Adding

**∠**

**EPD on both sides, we get**

**∠**

**EPA +**

**∠**

**EPD =**

**∠**

**DPB +**

**∠**

**EPD**

**⇒**

**∠**

**APD =**

**∠**

**BPE**

**(i)
Now, in ∆DAP and ∆EBP, we have
**

**∠**

**PAD =**

**∠**

**PBE [**

**∵∠**

**BAD =**

**∠**

**ABE]**

AP = BP [Proved above]

AP = BP [Proved above]

**∠**

**DPA =**

**∠**

**EPB [Proved above]**

**∴**

**∆DAP**

**≅**

**∆EBP [By ASA congruency]**

**(ii)
Since, ∆ DAP ****≅**** ∆ EBP
**

**⇒**

**AD = BE [By C.P.C.T.]**

**Ex 7.1 Class 9 Maths Question
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) ∆AMC

(i) ∆AMC

**≅**

**∆BMD**

(ii)

(ii)

**∠**

**DBC is a right angle**

(iii) ∆DBC

(iii) ∆DBC

**≅**

**∆ACB**

(iv) CM =

(iv) CM =

**AB**

Solution:

Since M is the mid – point of AB.

Solution:

Since M is the mid – point of AB.

**∴**

**BM = AM**

**(i) In
∆AMC and ∆BMD, we have
CM = DM [Given]
**

**∠**

**AMC =**

**∠**

**BMD [Vertically opposite angles]**

AM = BM [Proved above]

AM = BM [Proved above]

**∴**

**∆AMC**

**≅**

**∆BMD [By SAS congruency]**

**(ii)
Since ∆AMC ****≅**** ∆BMD
**

**⇒**

**∠**

**MAC =**

**∠**

**MBD [By C.P.C.T.]**

But they form a pair of alternate interior angles.

But they form a pair of alternate interior angles.

**∴**

**AC || DB**

Now, BC is a transversal which intersects parallel lines AC and DB,

Now, BC is a transversal which intersects parallel lines AC and DB,

**∴**

**∠**

**BCA +**

**∠**

**DBC = 180° [Co-interior angles]**

But

But

**∠**

**BCA = 90° [∆ABC is right angled at C]**

**∴**

**90° +**

**∠**

**DBC = 180°**

**⇒**

**∠**

**DBC = 90°**

**(iii)
Again, ∆AMC ****≅**** ∆BMD [Proved above]
**

**∴**

**AC = BD [By C.P.C.T.]**

Now, in ∆DBC and ∆ACB, we have

BD = CA [Proved above]

Now, in ∆DBC and ∆ACB, we have

BD = CA [Proved above]

**∠**

**DBC =**

**∠**

**ACB [Each 90°]**

BC = CB [Common]

BC = CB [Common]

**∴**

**∆DBC**

**≅**

**∆ACB [By SAS congruency]**

**(iv)
As ∆DBC ****≅**** ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
**

**∴**

**CM =**
DC =
AB

**⇒**

**CM =**
AB

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