RS Aggarwal Test: Triangles - Class 10 MCQ

The angle bisector theoremstates that an angle bisector divides the opposite side of a triangle into two segments that are proportional to the triangle's other two sides. In other words,
AB/BD = AC/CD.
Now
AB/AC=BD/DC
Which is the required ratio .
Thats how BD/DC=3/4

Δ ABC is an equilateral triangle

By Pythagoras theorem in triangle ABD

AB² = AD² + BD²

but BD = 1/2 BC (∵ In a triangle, the perpendicular from the vertex to the base bisects the base)

thus AB² = AD² + {1/2 BC}²

AB² = AD² + 1/4 BC²

4 AB² = 4AD² + BC²

4 AB² - BC² = 4 AD²

(as AB = BC we can subtract them )

Thus 3AB² = 4AD²

Given :△ABC, D, E and F are mid points of AB, BC, CA respectively.
Using mid point theorem we prove that □ADEF, □DBEF and □DECF are parallelograms. The diagonal of a parallelogram divides the parallelogram into two congruent triangles. So all triangles are congruent to each other. And each small triangle is similar to the original triangle.

Angles of ΔABC:

• ∠A = 50°
• ∠B = 70°
• ∠C = 60°

Angles of ΔDEF:

• ∠D = 60°
• ∠E = 70°
• ∠F = 50°

We can observe the following:

• ∠A = 50° corresponds to ∠F = 50°
• ∠B = 70° corresponds to ∠E = 70°
• ∠C = 60° corresponds to ∠D = 60°

So, the corresponding angles of ΔABC and ΔDEF are equal, meaning the two triangles are similar.

The correct answer is:

d) ΔFED.

In ΔCAB and ΔADB
Angle B is common and Angle A=Angle D
So the triangles are similar

a=cb/p
Now applying pythagoras theorem in 
Δ ABC
H² = P²+B²
BC²=AC²+AB²

Construct DG || BF
In triangle ADG,
AE = ED ( given)
EF || DG ( by construction)
a line passing through the mid point of any side of a triangle, parallel to the other side, bisects the third side
So AF = FG    ...(1)
Now in triangle CBF
BD = DC ( given)
DG||BF ( by construction)
a line passing through the mid point of any side of a triangle, parallel to the other side, bisects the third side
So, FG = GC …(2)
Now, by (1) & (2)
AF = FG = GC
 AF : FC = 1:2

since PQ = PR

so ∠Q = ∠R = 40^o

so ∠P comes out to be 100^o

and since PS is a median, so it'll bisect the ∠P into two equal parts

so, ∠QPS = 50^o