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ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
(i) ABE ≅ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle
(i) In ΔABE and ΔACF,
∠AEB = ∠AFC (Each 90º)
∠A = ∠A (Common angle)
BE = CF (Given)
∴ ΔABE ≅ ΔACF (By AAS congruence rule)
(ii) It has already been proved that
ΔABE ≅ ΔACF
⇒ AB = AC (By CPCT)
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠ A
(i) It is given that in triangle ABC, AB = AC
⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)
⇒ 
∠ACB = ∠ABC
⇒ ∠OCB = ∠OBC
⇒ OB = OC (Sides opposite to equal angles of a triangle are also equal)
(ii) In ΔOAB and ΔOAC,
AO =AO (Common)
AB = AC (Given)
OB = OC (Proved above)
Therefore, ΔOAB ≅ ΔOAC (By SSS congruence rule)
⇒ ∠BAO = ∠CAO (CPCT)
⇒ AO bisects ∠A.
In triangle ABC, AD is the perpendicular bisector of BC (see Figure). Show that triabgle ABC is an isosceles triangle in which AB = AC
In ΔADC and ΔADB,
AD = AD (Common)
∠ADC =∠ADB (Each 90º)
CD = BD (AD is the perpendicular bisector of BC)
∴ ΔADC ≅ ΔADB (By SAS congruence rule)
∴AB = AC (By CPCT)
Therefore, ABC is an isosceles triangle in which AB = AC.
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.
In ΔAEB and ΔAFC,
∠AEB and ∠AFC (Each 90º)
∠A = ∠A (Common angle)
AB = AC (Given)
∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)
⇒ BE = CF (By CPCT)
ABC and DBC are two isosceles triangles on the same base BC (see Figure). Show that
∠ ABD = ∠ ACD.
Let us join AD.
In ΔABD and ΔACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common side)
∴ ΔABD 
ΔACD (By SSS congruence rule)
⇒ ∠ABD = ∠ACD (By CPCT)
Triangle ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Figure). Show that ∠ BCD is a right angle.
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)
In ΔACD,
AC = AD
⇒ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)
In ΔBCD,
∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)
⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º
⇒ 2(∠ACB + ∠ACD) = 180º
⇒ 2(∠BCD) = 180º
⇒ ∠BCD = 90º
ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.
It is given that
AB = AC
⇒ ∠C = ∠B (Angles opposite to equal sides are also equal)
In ΔABC,
∠A + ∠B + ∠C = 180º (Angle sum property of a triangle)
⇒ 90º + ∠B + ∠C = 180º
⇒ 90º + ∠B + ∠B = 180º
⇒ 2 ∠B = 90º
⇒ ∠B = 45º
∴ ∠B = ∠C = 45º
Show that the angles of an equilateral triangle are 60° each.
Let us consider that ABC is an equilateral triangle.
Therefore, AB = BC = AC
AB = AC
⇒ ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)
Also,
AC = BC
⇒ ∠B = ∠A (Angles opposite to equal sides of a triangle are equal)
Therefore, we obtain
∠A = ∠B = ∠C
In ΔABC,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠A + ∠A = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
⇒ ∠A = ∠B = ∠C = 60°
Hence, in an equilateral triangle, all interior angles are of measure 60º.
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