Solution:
In Triangle ABD,∠BAD + ∠B + 90° = 180° or, ∠BAD + ∠B = 90° - - - - - - - (1) Now, in Triangle ABC, ∠ACB + ∠B + ∠A = 180° ∠ACB + ∠B = 180° - 90° ∠ACB = 90° - ∠B - - - - - (2) From (1) and (2), ∠BAD = ∠ACB So, AC = CD

Solution:
Since, ΔABC and ΔPQR are similar triangles. then, ∠B = ∠Q = 83° Thus, in ΔABC, ∠C = 180° - (∠A + ∠ B) or, ∠C = 180° - (47° + 83°) ∠C = 50°

Solution:
According to question,

Given :
AB = 10 cm
AD = 4 cm
DE || AC
ΔABC ∼ ΔDBE
$$\eqalign{ \therefore \frac{{BD}}{{AD}} = \frac{{BE}}{{CE}} \cr \,\,\,\,\,\,\frac{{BE}}{{CE}} = \frac{6}{4} \cr \,\,\,\,\,\,\frac{{BE}}{{CE}} = \frac{3}{2} \cr \,\,\,\,\,\,BE:CE = 3:2 \cr} $$

Solution:
According to question,
Let sides of the triangle be 3x, 4x, 6x
Now check the square of biggest side and sum of square of two smallest side and check which is greater.
∴ (3x)² + (4x)² < (6x)²
⇒ 25x² < 36x²
∴ The triangle will be obtuse angled triangle.

Solution:

Let BC = x and Ad = y, then as per bisector theorem,
$$\frac{{{\text{BD}}}}{{{\text{DC}}}} = \frac{{{\text{AB}}}}{{{\text{AC}}}} = \frac{4}{3}$$
Hence, BD = $$\frac{{4{\text{x}}}}{7}$$ and DC = $$\frac{{3{\text{x}}}}{7}$$
Now, in ΔABD using cosine rule,
$$\cos {30^ \circ } = \frac{{{4^2} + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$
$${\text{or,}}\,4\sqrt {3{\text{y}}} = {16 + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} } $$       - - - - - - (i)
Similarly in ΔADC,
$$\cos {30^ \circ } = \frac{{{3^2} + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$
$${\text{or,}}\,3\sqrt {3{\text{y}}} = {9 + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} } $$       - - - - - - - - (ii)
From equation (i) and (ii), we get
$${\text{y}} = \frac{{12\sqrt 3 }}{7}$$

Solution:
Since, AB and CD bisects each other at O, Hence, BC = AD = 6 cm.

Solution:

$$\eqalign{ \frac{{AP}}{{PB}} = \frac{{AQ}}{{QC}} \cr \frac{{AQ}}{{QC}} = \frac{3}{2} \cr AQ:QC = 3:2 \cr} $$