**Triangles**

**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 ∠A.

Now, In ∆ABC and ∆ABD,

AC = AD (Given)

∠ CAB = ∠ DAB ( AB bisects ∠ CAB)

and AB = AB (Common)

∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)

∴ BC = BD (By CPCT)

**Question 2.ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that**

(i) ∆ABD ≅ ∆BAC

(ii) BD = AC

(iii) ∠ABD = ∠ BAC

**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)

∴ ∆ 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.]

**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

∠BOC = ∠AOD

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.

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

∴ ∠BAC = ∠DCA …(1) [Alternate interior angles]

Also l || m and AC is a transversal,

∴ ∠BCA = ∠DAC …(2)

[Alternate interior angles]

Now, in ∆ABC and ∆CDA, we have

∠BAC = ∠DCA [From (1)]

CA = AC [Common]

∠BCA = ∠DAC [From (2)]

∴ ∆ABC ≅ ∆CDA [By ASA congruency]

**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 ≅ ∆AQB(ii) BP = BQ or B is equidistant from the arms ot ∠A.**

**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

∠ABP = ∠ABQ [Proved above]

AB = BA [Common]

∠PAB = ∠QAB [Given]

∴ ∆APB ≅ ∆AQB [By ASA congruency]

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 ∠A.

**Question 6.In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.**

**Solution:**We have, ∠BAD = ∠EAC

Adding ∠DAC on both sides, we have

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒ ∠BAC = ∠DAE

Now, in ∆ABC and ∆ADE. we have

∠BAC = ∠DAE [Proved above]

AB = AD [Given]

AC = AE [Given]

∴ ∆ABC ≅ ∆ADE [By SAS congruency]

⇒ BC = DE [By C.P.C.T.]

**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 ≅ ∆EBP(ii) AD = BE**

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

∴ AP = BP

∠EPA = ∠DPB [Given]

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]

∠DPA = ∠EPB [Proved above]

∴ ∆DAP ≅ ∆EBP [By ASA congruency]

**(ii) Since, ∆ DAP ≅ ∆ EBP**⇒ AD = BE [By C.P.C.T.]

**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 ≅ ∆BMD(ii) ∠DBC is a right angle(iii) ∆DBC ≅ ∆ACB(iv) CM = 12 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]

∴ ∆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.

∴ AC || DB

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

∴ ∠BCA + ∠DBC = 180° [Co-interior angles]

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]

∠DBC = ∠ACB [Each 90°]

BC = CB [Common]

∴ ∆DBC ≅ ∆ACB [By SAS congruency]

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

But DM = CM [Given]

∴ CM = 12DC = 12AB

⇒ CM = 12AB