Triangles

Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

Solution:
In ∆ABC and ∆ABD, we have
AC = AD (Given)
∠ CAB = ∠ DAB (∵ AB bisects ∠A)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that

Solution:
In ∆ ABC and ∆ BAC, we have
AD = BC (Given)
∠DAB = ∠CBA (Given)
and AB = AB (Common)
∴ ∆ ABD ≅ ∆BAC (By SAS congruence axiom)
Hence, BD = AC (By CPCT)
and ∠ABD= ∠BAC (By CPCT)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.

Solution
From figure, we have
∠ 1 = ∠ 2 (Vertically opposite angles).. .(i)
∠ 1 = ∠ 6 (Corresponding angles).(ii)
∠ 6 = ∠ 4 (Corresponding angles) .(iii)
From Eqs. (i) (ii) and (iii), we have
∠ 1 = ∠ 4
and ∠ 2 = ∠ 4 ...(iv)

In ∆ABC and ∆CDA, we have ∠ 4 = ∠ 2 [From Eq. (iv] ∠5 = ∠ 3 (Alternate interior angles) and AC = AC (Common) ∴ ∆ABC ≅ ∆ CDA (By AAS congruence axiom)

Question 5.
Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ?A.

Solution

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

Solution

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that
(i) OB = OC
(ii) AO bisects ∠A

Solution: (i) In ∆ ABC, we have AB = AC (Given) ⇒ ∠B = ∠C (∵ Angles opposite to equal sides are equal)

Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

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.

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) ∆ABE = ∆ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.

Question 5.
ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD.

Question 6.
∆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.

Solution

Question 8.
Show that the angles of an equilateral triangle are 60° each.

Solution:
Let ∆ ABC be an equilateral triangle, such that

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that

• (i) ∆ABD = ∆ACD
• (ii) ∆ABP = ∆ACP
• (iii) AP bisects ∠A as well as ∠D
• (iv) AP is the perpendicular bisector of BC.

Solution:
Given ∆ABC and ∆DBC are two isosceles triangles having common
base BC, suchthat AB=AC and DB=OC.

To prove:

• (i) ∆ABD = ∆ACD
• (ii) ∆ABP = ∆ACP
• (iii) AP bisects ∠A as well as ∠D
• (iv) AP is the perpendicular bisector of BC.

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A

Solution: In ∆ ABD and ∆ ACD, we have

Solution

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution: In ∆BEC and ∆CFB, we have

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.

Solution: Let ABC be a right angled triangle, such that ∠ ABC = 90°

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
 

Solution:

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.

Solution:
Suppose OM and ON be the perpendicular bisectors of sides BC and AC of ∆ ABC.

So, O is equidistant from two endpoints 0 and C of line segment BC as O lies on the perpendicular bisector of BC. Similarly, O is equidistant from C and A Hence, O be an orthocentre of ∆ABC.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Solution:
Suppose BN and CM be the bisectors of ∠ ABC and ∠ ACB, respectively intersect AC and AB at N and M, respectively.

Since, O lies on the bisector BN of ∠ ABC, so O will be equidistant from BA and BC. Again, O lies on the bisector CM of ∠ ACB.

So, O will be equidistant from CA and BC. Thus, O will be equidistant from AB, BC and CA Hence, O be a circumcentre of ∆ABC.

Question 3.
In a huge park, people are concentrated at three points (see figure)
 

• A: where these are different slides and swings for children.
• B: near which a man-made lake is situated.
• C: which is near to a large parking and exist.
   

Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?
[Hint The parlour should be equidistant from A, B and C.]

Solution:
The ice-cream parlor should be equidistant from A B and C for which the point of intersection of perpendicular bisectors of AB, BC, and CA should be situated.
So, O is the required point which is equidistant from A B and C.

Question 4.
Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Solution:
We first divide the hexagon into six equilateral triangles of side 5cm as follow.

We take one triangle from six equilateral triangles as shown above and make as many equilateral triangles of one side 1 cm as shown in the figure.

The number of equilateral triangles of side 1 cm = 1 + 3 + 5 + 7 + 9 = 25
So, the total number of triangles in the hexagon = 6x 25 = 150
To find the number of triangles in the Fig. (ii), we adopt the same procedure.
So, the number of triangles in the Fig. (ii) = 12 x 25 = 30Q Hence, Fig. (ii) has more triangles.