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**Question 1. In quadrilateral ACBD, AC = AD and AB bisects âˆ A (see figure). Show that Î”ABC â‰Œ Î”ABD. What can you say about BC and BD? Solution:** In quadrilateral ABCD we have AC = AD

and AB being the bisector of âˆ A.

Now, in Î”ABC and Î”ABD,

AC = AD [Given]

AB = AB [Common]

âˆ CAB = âˆ DAB [âˆµ AB bisects âˆ CAD]

âˆ´ Using SAS criteria, we have Î”ABC â‰Œ Î”ABD.

âˆµ Corresponding parts of congruent triangles (c.p.c.t) are equal.

âˆ´ BC = BD.

**Question 2. ABCD is a quadrilateral in which AD = BC and âˆ DAB = âˆ CBA (see Figure). Prove that (i) **Î”

Solution:

In Î”ABD and Î”BAC,

AD = BC [Given]

AB = BA [Common]

âˆ DAB = âˆ CBA [Given]

âˆ´ Using SAS criteria, we have Î”ABD â‰Œ Î”BAC

(ii) âˆµ Î”ABD â‰Œ Î”BAC

âˆ´ Their corresponding parts are equal.

â‡’ BD = AC

(iii) Since Î”ABD â‰Œ Î”BAC

âˆ´ Their corresponding parts are equal.

â‡’ âˆ ABD = âˆ BAC.

**Question 3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB. Solution:** We have âˆ ABC = 90Â° and âˆ BAD = 90Â°

Also AB and CD intersect at O.

âˆ´ Vertically opposite angles are equal.

Now, in Î”OBC and Î”OAD, we have

âˆ ABC = âˆ BAD [each = 90Â°]

BC = AD [Given]

âˆ BOC = âˆ AOD [vertically opposite angles]

âˆ´ Using ASA criteria, we have Î”OBC â‰Œ Î”OAD

â‡’ OB = OA [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 **Î”**ABC â‰Œ **Î”**CDA.**

**Solution:** âˆµ â„“ || m and AC is a transversal,

âˆ´ âˆ BAC = âˆ DCA [Alternate interior angles]

Also p || q and AC is a transversal,

âˆ´ âˆ BCA = âˆ DAC [Alternate interior angles]

Now, in Î”ABC and Î”CDA,

âˆ BAC = âˆ DCA [Proved]

âˆ BCA = âˆ DAC [Proved]

CA = AC [Common]

âˆ´ Using ASA criteria, we have Î”ABC â‰Œ Î”CDA

**Question 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 âˆ A (see Figure). Show that: (i) **Î”

(ii) BP = BQ or B is equidistant from the arms of âˆ A.

**Solution:** We have, â„“ as the bisector of QAP.

âˆ´ âˆ QAB = âˆ PAB

âˆ Q= âˆ P [each = 90Â°]

â‡’ Third âˆ ABQ = Third âˆ ABP (i)

Now, in Î”APB and Î”AQB, we have

AB = AB [common]

âˆ ABP = âˆ ABQ [proved]

âˆ PAB = âˆ QAB [proved]

âˆ´ Using SAS criteria, we have Î”APB â‰Œ Î”AQB

(ii) Since Î”APB â‰Œ Î”AQB

âˆ´ Their corresponding angles are equal.

â‡’ BP = BQ

i.e. [Perpendicular distance of B from AP] = [Perpendicular distance of B from AQ]

Thus, the point B is equidistant from the arms of âˆ A.

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

**Solution:** We have âˆ BAD = âˆ EAC

Adding âˆ DAC on both sides, we have

âˆ BAD + âˆ DAC = âˆ EAC + âˆ DAC

â‡’ âˆ BAC = âˆ DAE

Now, in Î”ABC and Î”ADE, we have

âˆ BAC = âˆ DAE [Proved]

AB = AD [Given]

AC = AE [Given]

âˆ´ Î”ABC â‰Œ Î”ADE [Using SAS criteria]

Since Î”ABC â‰Œ Î”ADE, therefore, their corresponding parts are equal.

â‡’ BC = DE.

**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 âˆ BAD = âˆ ABE and âˆ EPA = âˆ DPB (see figure). Show that (i) **Î”**DAP â‰Œ **Î”**EBP (ii) AD = BE**

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

âˆ´ AP = BP

âˆ EPA = âˆ DPB [Given]

Adding âˆ EPD on both sides, we get

âˆ EPA + âˆ EPD = âˆ DPB + âˆ EPD

â‡’ APD = âˆ BPE

(i) Now, in Î”DAP â‰Œ Î”EBP, we have

AP = BP. [Proved]

âˆ PAD = âˆ PBE [âˆµ It is given that âˆ BAD = âˆ ABE]

âˆ DPA = âˆ EPB [Proved]

âˆ´ Using ASA criteria, we have

Î”DAP â‰ŒÎ”EBP

(ii) Since, Î”DAP â‰ŒÎ”EBP

âˆ´ Their corresponding parts are equal. â‡’ AD = BE.

**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) **Î”**AMC â‰Œ **Î”**BMD (ii) âˆ DBC is a right angle. (iii) **Î”**DBC â‰Œ **Î”**ACB (iv) CM = (1/2)AB Solution:** âˆµ M is the mid-point of AB.

âˆ´ BM = AM [Given]

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

AM = BM [Proved]

âˆ AMC = âˆ BMD [Vertically opposite angles]

âˆ´ Î”AMC â‰Œ Î”BMD (SAS criteria)

(ii) âˆµ Î”AMC â‰Œ Î”BMD

âˆ´ Their corresponding parts are equal.

â‡’ âˆ MAC = âˆ MBD But they form a pair of alternate interior angles.

âˆ´ AC || DB

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

âˆ´ âˆ BCA + âˆ DBC = 180Â°

But âˆ BCA = 90 [âˆµ Î”ABC is right angled at C]

âˆ´ 90Â° + âˆ DBC = 180Â°

or âˆ DBC = 180Â° âˆ 90Â° = 90Â°

Thus, âˆ DBC = 90Â°

(iii) Again, Î”AMC â‰Œ Î”BMD [Proved]

âˆ´ AC = BD [c.p.c.t]

Now, in Î”DBC and Î”ACB, we have

âˆ DBC = âˆ ACB [Each = 90Â°]

BD = CA [Proved]

BC = CB [Common]

âˆ´ Using SAS criteria, we have Î”DBC â‰Œ Î”ACB

(iv) âˆµ Î”DBC â‰Œ Î”ACB

âˆ´ Their corresponding parts are equal.

â‡’ DC = AB

But DM = CM [Given]

âˆ´ CM = (1/2) DC = (1/2) AB

â‡’ CM = (1/2) AB

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