Question 1.
In $\Delta ABC$ , ray AD bisects $ \angle A$ and intersects BC in D. If BC = a, AC = b and AB = c, prove that:
i. $BD = \frac {ac}{b+c}$
ii.$DC = \frac {ab}{c+b}$ Solution

Now,
$CD = BC -BD$
$CD= a - \frac {ac}{b+c}$
$DC = \frac {ab}{b+c}$

Question 2.
Two poles of heights a and b metres are standing vertically on a level ground r metres apart. Prove that the height c of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by $\frac {ab}{a+b}$ i.e $c = \frac {ab}{a+b}$ Solution

Let FB=x and FD=y
Now in $\Delta ABD$ and $\Delta EFD$
$\angle ABD= \angle EFD$
$\angle ADB= \angle EDF$ (common angle)
By SS criteria,
$\Delta ABD \sim \Delta EFD$
$\frac {AB}{EF} = \frac {BD}{FD}$
$\frac {a}{c} = \frac {x+y}{y}$
$\frac {a}{c} -1 = \frac {x+y}{y} -1$
$\frac {a-c}{c}= \frac {x}{y}$ -(1)

Question 3.
D and E are points on the sides CA and CB respectively of $\Delta ABC$ right-angled at C. Prove that $AE^2 + BD^2 = AB^2 + DE^2$ Solution

This can be easily solved using Pythagorus theorem
In Right angle triangle ACE,By Pythagoras Theorem
$AC^2+ CE^2 = AE^2$ --(1)
In Right angle triangle DBC,By Pythagoras Theorem
$DC^2+ BC^2 = BD^2$ -(2)
In Right angle triangle ABC,By Pythagoras Theorem
$AC^2 + BC^2 = AB^2$ ---(3)

In Right angle triangle DEC,By Pythagoras Theorem
$DC^2 + CE^2 = DE^2$ ---(4)

Question 4.
In a quadrilateral ABCD, $\angle B = 90$ , $AD^2= AB^2 + BC^2 + CD^2$, prove that $\angle ACD = 90$ Solution

It is given $\angle B = 90$
In right angle triangle ABC
$AC^2 = AB^2 + BC^2$ -(1)
Now it is given
$AD^2= AB^2 + BC^2 + CD^2$
From (1)
$AD^2 = AC^2 + CD^2$
By converse of Pythagoras theorem
$\angle ACD = 90$

Question 5.
ABC is a triangle in which AB = AC and D is a point on AC such that
$BC^2 =AC \times CD$ . Prove that BD=BC Solution

Now given
$BC^2 =AC \times CD$
$ \frac {BC}{AC} = \frac {CD}{BC}$

In $\Delta ABC$ and $\Delta BDC$
$ \frac {BC}{AC} = \frac {CD}{BC}$
$\angle ACB = \angle BCD$
Hence
$\Delta ABC \sim \Delta BDC$
By CPCT
$\frac {BD}{BC} = \frac {AB}{AC}$
Now AB =AC
$\frac {BD}{BC}=1$
BD=BC

Question 6.
if $ \Delta ABC \sim \Delta DEF$ and their sides are of lengths (in cm) as marked along them, then find the lengths of the sides of each triangle.

Given
$ \Delta ABC \sim \Delta DEF$
Therefore,
$\frac {AB}{DE} =\frac {BC}{EF}=\frac {CA}{FD}$
So,
$\frac {2x-1}{18}=\frac {2x+2}{3x+9}= \frac {3x}{6x}$
$\frac {2x-1}{18}=\frac {2x+2}{3x+9}$
or
$x = 5$
Therefore,
AB = 2 * 5 –1= 9, BC = 2 * 5 + 2 = 12,
CA = 3 * 5 = 15, DE = 18, EF = 3* 5 + 9 = 24 and FD = 6 * 5 = 30
Hence, AB = 9 cm, BC = 12 cm, CA = 15 cm
DE = 18 cm, EF = 24 cm and FD = 30 cm

Question 7. True and False statement
(a) All quadrilateral are similar.
(b) All circles are similar.
(c ) All isosceles triangles are similar.
(d) All 30Â°, 60Â°, 90Â° triangles are similar. Solution

a. False
b. True
c. False
d. True

Question 8.
In the below figure PA, QB and RC are each perpendicular to AC.
Prove that
$ \frac {1}{x} + \frac {1}{y} = \frac {1}{z}$

This question is exactly similar to Question 2. Please check the solution

Question 9.
ABC is a right triangle right angled at C. Let BC = a, CA = b AB = c and let p be the length of perpendicular from C on AB, prove that
(i) $cp = ab$
(ii) $ \frac {1}{p^2}= \frac {1}{a^2}+ \frac {1}{b^2}$ Solution

Area of the Triangle ABC
$=\frac {1}{2} \times base \times height$
Now
Area of the Triangle ABC=$\frac {1}{2} \times a \times b= \frac {ab}{2}$
Now it can be also calculated as
Area of the Triangle ABC=$\frac {1}{2} \times c \times p= \frac {pc}{2}$
Since area is equal
$\frac {ab}{2}=\frac {pc}{2}$
$cp =ab$ -(1)

Now in right angle triangle ABC
$AC^2 = AB^2 + BC^2$
$c^2 = a^2 + b^2$
From (1)
$( \frac {ab}{p})^2 = a^2 + b^2$
$\frac {a^2b^2}{p^2} = a^2 + b^2$
Dividing by $a^2b^2$
$ \frac {1}{p^2}= \frac {1}{a^2}+ \frac {1}{b^2}$

Question 10.
Diagonals of a trapezium ABCD with AB||DC intersect each other at the point O. If AB = 2DC, find ratio of the areas of AOB and COD Solution

In $\Delta AOB$ and $\Delta COD$, we have
$\angle AOB= \angle COD$ ( Vertically Opposite angles)
$\angle OCD= \angle OAB$ ( Alternate angles)
$\angle ODC=\angle OBA$ ( Alternate angles)
by AAA similarity
$\Delta AOB \sim \Delta COD$
Now Area of Similar Triangle is given by
$\frac {Area AOB}{Area COD} = \frac {AB^2}{CD^2}$
Now AB= 2CD
Hence
$\frac {Area AOB}{Area COD} = \frac {4}{1}$