Congurency In Triangles

Exercise 7.2

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.

congurence trinagles 2

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)
So, `∠BAO=∠CAO`
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.

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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.

congurence trinagles 6

Answer: In ΔABE and ΔACF
`AB = AC` (given)
`∠BAE=∠CAF` (common to both triangles)
`∠CFA=∠BEQ` (right angles)
So, `ΔABE≅ ΔACF` (ASA rule)
So, `BE=CF`

Question 4: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that

  1. `ΔABE≅ ΔACF`
  2. AB = AC, i.e., ABC is an isosceles triangle.

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.

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Answer: `∠ABC=∠ACB`
`∠DBC=∠DCB`
So, `∠ABC+∠DBC=∠ACB+∠DCB`
Or, `∠ABD=∠ACD`

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.

congurence trinagles 10

Answer: In ΔADC and ΔABC
`AD=AB`
`AC=AC`
`∠ACB=∠ABC`
`∠ACD=∠ACD`
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
`180°=360°-2∠ACB-2∠ACD`
Or, `180°=360°-2(∠ACB+∠ACD)`
Or, `2(∠ACB+∠ACD)=180°`
Or, `∠ACB+∠ACD=∠BCD=90°`

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°.
So, `∠A+∠B+∠C=190°`
Or, `90°+∠B+∠C=180°`
Or, `∠B+∠C=180°-90°=90°`
Or, `∠B=∠C=45°`

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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