Answer: Option B. -> Equilateral
a2 + b2 + c2 = ab + cb + ca
This is true only when a = b = c
So, triangle will be equilateral

Answer: Option C. -> 10 cm
$$\eqalign{
& \vartriangle ADE \simeq \vartriangle ABC \cr
& \frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} \cr
& \frac{{1.5}}{{7.5}} = \frac{2}{{BC}} \cr
& BC = 10\,{\text{cm}} \cr} $$

Answer: Option B. -> 100°
∵ Sum of all angles of a triangle = 180°
So, ∠BAC = 180 - (90 + 70)
∠BAC = 20°
So, ∠BIC = 90 + $$\frac{1}{2}$$ ∠A
∠BIC = 90° + $$\frac{1}{2}$$ × 20°
∠BIC = 100°

Answer: Option B. -> 5cm, 8cm, 15cm
For a triangle sum of 2 sides is always greater than the third side.
Hence, combination (5, 8, 15) never be possible

Answer: Option B. -> 6 cm
∵ ΔPQR ∼ ΔPMN
∵ ΔPQR is equilateral
∴ PQ = PR = QR
So, ΔPMN must be equilateral
So, MN = PN = 6 cm

Answer: Option C. -> Right-Angled
According to question,
Sides AB = x2 - 1
          BC = 2x
          AC = x2 + 1
By using Pythagoras theorem
AC2 = AB2 + BC2
(x2 + 1)2 = (x2 - 1)2 + (2x)2
x4 + 1 + 2x2 = x2 + 1 - 2x2 + 4x2
(x2 + 1)2 = (x2 + 1)2
∴ The triangle is right angle Δ

Answer: Option A. -> AX = BY = CZ
According to question,
In an equilateral triangle
AB = BC = AC
∠A = ∠B = ∠C = 60°
∴ AX = BY = CZ
(All altitudes are same in an equilateral triangles)

Answer: Option B. -> 108°
According to question,
Let ∠CAB = x and ∠CBA = y
$$\eqalign{
& \Rightarrow \angle CAD = \frac{{180 - x}}{2} = 90 - \frac{x}{2} \cr
& {\text{and}} \cr
& \Rightarrow \angle EBC = \frac{{180 - y}}{2} = 90 - \frac{y}{2} \cr
& {\text{also}}\,\angle AEB = \angle EAB = x \cr} $$
(∵ AB = EB ⇒ ABE is an isosceles triangle)
and ∠ADB = ∠ABD = y
(∵ AB = AD ⇒ ADB is an isosceles triangle)
In ΔAEB,
∠AEB + ∠ABE + ∠BAE = 180°
x + x + y + 90 - $$\frac{y}{2}$$ = 180°
⇒ 4x + y = 180°
Similarly in ΔADB
4y + x = 180°
⇒ 4y + x + 4x + y = 180 + 180
⇒ 5x + 5y = 360°
⇒ x + y = 72°
In triangle ABC,
∠ACB + x + y = 180°
⇒ ∠ACB = 180 - 72
⇒ ∠ACB = 108°

Answer: Option D. -> 3 : 2
According to question,
Given :
AB = 10 cm
AD = 4 cm
DE || AC
ΔABC ∼ ΔDBE
$$\eqalign{
& \therefore \frac{{BD}}{{AD}} = \frac{{BE}}{{CE}} \cr
& \,\,\,\,\,\,\frac{{BE}}{{CE}} = \frac{6}{4} \cr
& \,\,\,\,\,\,\frac{{BE}}{{CE}} = \frac{3}{2} \cr
& \,\,\,\,\,\,BE:CE = 3:2 \cr} $$

Answer: Option B. -> Equilateral
According to question,
 $$ \Rightarrow \left( {x + {{15}^ \circ }} \right) + {\left( {\frac{{6x}}{5} + 6} \right)^ \circ } + $$      $${\left( {\frac{{2x}}{3} + 30} \right)^ \circ } = $$    $${180^ \circ }$$
 $$\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\angle A + \angle B + \angle C = {{180}^ \circ }} \right\}$$
 $$ \Rightarrow x + \frac{{6x}}{5} + \frac{{2x}}{3} = $$     $${180^ \circ } - \left(15 + 6 + 30\right)$$
$$\eqalign{
& \Rightarrow \frac{{15x + 18x + 10x}}{{15}} = 180 - 51 \cr
& \Rightarrow 43x = 129 \times 15 \cr
& \Rightarrow x = {45^ \circ } \cr
& \Rightarrow {\text{each}}\,{\text{angle}} \cr
& \Rightarrow {\left( {x + 15} \right)^ \circ } = 45 + 15 = {60^ \circ } \cr
& \Rightarrow {\left( {\frac{{6x}}{5} + 6} \right)^ \circ } = {60^ \circ } \cr
& \Rightarrow {\left( {\frac{{2x}}{3} + 30} \right)^ \circ } = {60^ \circ } \cr} $$
∵ All three angles are equal 60°
⇒ Triangle will be equilateral triangle