Question 1
Question: Prove that if a line be drawn Parallel to the base of aD , the portion intercepted by the sides is bisected by the median to the base. [3 Mark]
Answer: 
Given:
In D ABC, PQ||BC. AD is the median of BC and meets BC in D.
To prove:
PR = RQ
R is midpoint of PQ.
Proof:
Since PQ||BC,
D APR and D ABD are equiangular.
D APR ~ DABD (AAA)
....(1)
Similarly, D ARQ ~ D ADC
....(2)
(From (1) and (2) both are equal to
)
Question 2
Question: In 
[3 Mark]
Answer: 
Given:
or
Proof:
In D ADC and D BDA,
(given)
(SAS similarity criterion)
The two triangles are equiangular.
....(1)
Now in D ABC,
(Angle sum property of a triangle)
(From (1))
Question 3
Question: ABC is a triangle in which AB = AC and D is a point on the side AC such that BC2 = AC x CD. Prove that BD = BC. [2 Mark]
Answer: 
Given:
AB = AC and BC2 = AC x CD
i.e., 
To prove: BD = BC In D BCA and D DCB,
| Note: |
| That we name triangle ABC as BCA to keep up the correspondence with D DCB. The corresponding angles at the vertices C in both triangles are equal. |
Proof:
(given)
(common angle)
(SAS similarity criterion)
....(1)But AB = AC (given)
BC = BD (FromNote:That we name triangle ABC as BCA to keep up the correspondence with D DCB. The corresponding angles at the vertices C in both triangles are equal.Question 4
Question: In the given figure, D ABC ~ D DEF. If AB = 2DE and area of D ABC = 56 sq.cm, then find the area of D DEF. [2 Mark]
Answer: We know that, if D ABC ~ D DEF, then
[The ratio of areas of two similar 
is equal to the ratio of the squares of their corresponding sides]
\ Area of D DEF = 14 sq.cm
Question 5
Question: In the given figure, DE ll BC and AD:DB = 5:4. find 
[2 Mark]
Answer: 
(By AA similarity criterion)
....(1)
In D DFE and D CFB,
(Alternate angle)
(Vertically opposite angle)
(AA Similarity)
Question 6
Question: D, E and F are the mid-point of the sides BC, CA and AB respectively of a D ABC. Determine the ratio of the areas of D DEF and D ABC. [3 Mar />
Answer: By mid-point theorem DE||BC
....(1)
AFDE is a parallelogram. (Since opposite sides are parallel)
(Opposite angles of a parallelogram are equal0
Similarly 
are equiangular and hence they are similar.
Question 7
Question: ABC is a triangle and DE is a line segment meeting AB in D and AC in E such that DE||BC and divides D ABC into two parts of equal area. Find
[3 Mark]
Answer: 
Given:
Area (D ADE) = Area (trapezium DECB) ... (1)
Area (D ABC) = Area (D ADE) + Area (trap DECB)
= Area (D ADE) + Area (D ADE)
Area (D ABC) = 2(Area (D ADE)) ... (2)
(By AA similarity criterion)
But 
(From (2))
Question 8
Question: In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, then prove that
[3 Mark]
Answer: Construction:
Draw AE
BC and DF
BC
Proof:
....(1)
Now in D AOE and D DOF, we have
(Vertically opposite angles)
(each angle = 90o)
(AA similarity)
....(2)
From (1) and (2), we get
| Note: |
| If two triangles have a common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases. |
The triangles ABC, ABD, ADC, AEC have a common vertex A and their bases are along the same line BC.
Then, we have
i.
ii.
iii.
The proof of (i) is shown below
Note:If two triangles have a common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases.Question 9
Question: In the figure, ABCD is a parallelogram. P is a point on BC such that BP:PC = 1:2. DP produced meets AB produced at Q. If area of D CPQ = 20 cm2, find [3 Mark]
i. Area of D BPQ
ii. Area of D CDP
iii. Area of parallelogram ABCD.
Answer: D CPQ and D BPQ have the common vertex Q, bases on the same line, therefore
... (1)
ii. 
(AAA Similarity)
(From 1)
= 4 x 10
= 40cm2
iii. 
= 40 + 20 = 60cm2
Area of (parallelogram ABCD) = 2 x Area (D QDC) = 120 cm2 (
If a parallelogram and a triangle are on the same base and between the same parallels; area of the parallelogram is twice the area of the triangle)
Question 10
Question: In a triangle, AD^ BC, prove that AB2- BD2= AC2- CD2. [2 Mark]
Answer: 
Given:
In D ADB,
(Pythagoras Theorem)
... (1)
In D ADC,
(Pythagoras theorem)
...(2)
L.H.S of (1) and (2) are equal.
\ R.H.S of (1) and (2) must be equal.
