Question 1: 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: (a) OB = OC (b) AO bisects ∠ A.
Answer: In ΔOBC
`∠OBC = ∠ OAC` (They are halves of angless B and C)
So, `OB = OC` (Sides opposite to equal angles)
In ΔAOB and ΔAOC
`AB = AC` (given)
`OB = OC` (proven earlier)
`∠ABO = ∠ACO` (they are halves of angles B and C)
So, `ΔAOB ≅ ΔAOC`(SAS rule)
It means that AO bisects ∠A
Question 2: In Δ ABC, AD is the perpendicular bisector of BC. Show that Δ ABC is an isosceles triangle in which AB = AC.
Answer: In ΔABD and ΔACD
`AD = AD` (common side)
`BD = CD` (given)
`∠ADB=∠ADC` (right angle)
So, `ΔABD≅ ΔACD`
So, `AB = AC`
This proves that ΔABC is isosceles triangle.
Question 3: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.
Answer: In ΔABE and ΔACF
`AB = AC` (given)
`∠BAE=∠CAF` (common to both triangles)
`∠CFA=∠BEQ` (right angles)
So, `ΔABE≅ ΔACF` (ASA rule)
Question 4: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that
Answer: This can be solved like previous question.
Question 5: ABC and DBC are two isosceles triangles on the same base BC. Show that angle ABD = angleACD.
Question 6: ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that angle BCD is a right angle.
Answer: In ΔADC and ΔABC
In ΔABC, `∠ACB+∠ABC+∠CAB=180°`
Or, `∠CAB=180°-2∠ACB` ----- (1)
Similarly, in ΔADC
`∠DAC=180°-2∠ACD` ------ (2)
As BD is a straight line, so `∠CAB+∠DAC=180°`
So, adding equations (1) and (2) we get
Question 7: ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Answer: If AB = AC then angles opposite to these sides will be equal. As you know the sum of all angles of a triangle is equal to 180°.
Question 8: Show that the angles of an equilateral triangle are 60° each.
Answer: As angles opposite to equal sides of a triangle are always equal. So, in case of equilateral triangle all angles will be equal. So they will measure one third of 180°, which is equal to 60°
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