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**Question 1. In an isosceles triangle ABC, with AB = AC, the bisectors of âˆ B and âˆ C intersect each other at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects âˆ A Solution: **(i) In Î”ABC, we have AB = AC [Given]

âˆ´ âˆ C= âˆ B

[Angle opposite to equal sides are equal]

â‡’ (1/2)âˆ C= (1/2)âˆ B

or âˆ OCB = âˆ OBC

â‡’ OB = OC [Sides opposite to equal angles are equal]

(ii) In Î”ABO and Î”ACO, we have

AB = AC [Given]

OB = OC [Proved]

âˆ OBA = âˆ OCA

âˆ´ Using SAS criteria,

Î”ABO â‰Œ Î”ACO

â‡’ âˆ OAB = âˆ OAC [c.p.c.t.]

â‡’ AO bisects âˆ A.

**Question 2. In **Î”**ABC, AD is the perpendicular bisector of BC (see figure). Show that **Î”**ABC is an isosceles triangle in which AB = AC. Solution:** âˆµ AD is bisector of BC.

âˆ´ BD = CD

Now, in Î”ABD and Î”ACD, we have:

AD = AD [Common]

âˆ ADB = âˆ ADC = 90Â° [âˆµ AD âŠ¥ BC]

BD = CD [Proved]

âˆ´ Î”ABD â‰Œ Î”ACD [SAS criteria]

âˆ´ Their corresponding parts are equal.

â‡’ AB = AC Thus, Î”ABC is an isosceles triangle.

**Question 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Figure) Show that these altitudes are equal. OR ABC is an isosceles triangle with AB = AC. Prove that the altitudes BE and CF of the triangle are equal.**

**Solution: **DABC is an isosceles triangle.

âˆ´ AB = AC

â‡’ âˆ ACB = âˆ ABC [âˆµ Angles opposite to equal sides are equal]

Now, in Î”BEC and Î”CFB, we have

âˆ EBC = âˆ FCB [Proved]

BC = CB [Common]

and âˆ BEC = âˆ CFB [Each = 90Â°]

âˆ´ Î”BEC â‰Œ Î”CFB [Using ASA criteria]

â‡’ Their corresponding parts are equal. i.e. BE = CF

**Question 4. ABC is a triangle is which altitudes BE and CF to sides AC and AB are equal (see figure). Show that (i) **Î”

(ii) AB = AC, i.e. ABC is an isosceles triangle.

Solution:

âˆ AEB = âˆ AFC

[each = 90Â° âˆµ BE âŠ¥ AC and CF âŠ¥ AB]

âˆ A= âˆ A [Common]

BE = CF [Given]

âˆ´ Î”ABE â‰Œ Î”ACF [Using AAS criterion]

(ii) Since, Î”ABE â‰Œ Î”ACF

âˆ´ Their corresponding parts are equal.

â‡’ AB = AC** **

**Question 5. ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that âˆ ABD = âˆ ACD. Solution:** In Î”ABC, we have AB = AC

[âˆµ DABC is an isosceles triangle]

But angles opposite to equal sides are equal.

âˆ´ âˆ ABC = âˆ ACB ...(1)

Again, in Î”BDC, we have BD = CD

[âˆµ Î”BDC is an isosceles triangle.]

âˆ´ âˆ CBD = âˆ BCD ...(2)

[Angles opposite to equal sides are equal]

Adding (1) and (2), we have

âˆ ABC + âˆ CBD = âˆ ACB + âˆ BCD

â‡’ âˆ ABD = âˆ ACD

**Question 6. **Î”**ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that âˆ BCD is a right angle. Solution: **âˆµ In Î”ABC, AB = AC [Given] â€¦(1)

AB = AD [Given] â€¦(2)

From (1) and (2), we have AC = AD

Now, in Î”ABC, we have âˆ B + âˆ ACB + âˆ BAC = 180Â°

â‡’ 2âˆ ACB + âˆ BAC = 180Â° â€¦(3)

[âˆµ âˆ B = âˆ ACB (Angles opposite to equal sides)]

In Î”ACD, âˆ D + âˆ ACD + âˆ CAD = 180Â°

â‡’ 2âˆ ACD + âˆ CAD = 180Â° ...(4)

[âˆµ âˆ D = âˆ ACD (angles opposite to equal sides)]

Adding (3) and (4), we have

2âˆ ACB + âˆ BAC + 2âˆ ACD + âˆ CAD = 180Â° + 180Â°

â‡’ 2[âˆ ACB + âˆ ACD] + [âˆ BAC + âˆ CAD] = 360Â°

â‡’ 2[âˆ BCD] + [180Â°] = 360Â° [âˆ BAC and âˆ CAD form a linear pair]

â‡’ 2âˆ BCD = 360Â° âˆ 180Â° = 180Âº

â‡’ âˆ BCD = (180^{0}/2)= 90Â°

Thus, âˆ BCD = 90Â°.

**Question 7. ABC is a right angled triangle in which âˆ A = 90Â° and AB = AC. Find âˆ B and âˆ C. Solution**: In Î”ABC, we have AB = AC [Given]

âˆ´ Their opposite angles are equal.

â‡’ âˆ ACB = âˆ ABC

Now, âˆ A + âˆ B + âˆ C = 180Â°

â‡’ 90Â° + âˆ B + âˆ C = 180Â° [âˆµ âˆ A = 90Â° (Given)]

â‡’ âˆ B + âˆ C = 180Â°

But âˆ ABC = âˆ ACB,

i.e. âˆ B = âˆ C

âˆ´ âˆ B= âˆ C = (90^{0}/2) = 45Â°

Thus, âˆ B = 45Â° and âˆ C = 45Â°

**Question 8. Show that the angles of an equilateral triangle are 60Â° each. Solution: **In Î”ABC, we have

AB = BC = CA [âˆµ ABC is an equilateral triangle]

âˆ´ AB = BC â‡’ âˆ A = âˆ C â€¦(1)

[âˆµ Angle opposite to equal sides are equal.]

Similarly, AC = BC â‡’ âˆ A = âˆ B

From (1) and (2), we have âˆ A= âˆ B = âˆ C

Let âˆ A= âˆ B = âˆ C = x

Since, âˆ A + âˆ B + âˆ C = 180Â°

âˆ´ x + x + x = 180Â°

â‡’ 3x = 180Â° or x =(180^{o}/3) = 60Â°

âˆ´ âˆ A= âˆ B = âˆ C = 60Â°

Thus, the angles of an equilateral triangle are 60Â° each.

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