Question 1:
In the given figure, if x = y and AB = CB, then prove that AE = CD.
Solution:
Given: AB = BC
and, xo = yo
To prove: AE = CD
Proof: In \(\Delta\)
Exterior \(\angle\)
\(\Rightarrow\)
Again, in \(\Delta\)
xo = \(\angle\)
Since, xo = yo [Given]
So, \(\angle\)
\(\Rightarrow \)
Thus in \(\Delta\)
\(\angle\)
BC = AB [Given]
and, \)
Thus by Angle-Side-Angle criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal.
So, CD = AE [Proved]
Question 2:
ABC is a triangle in which AB = AC. If the bisectors of \(\angle\)
Solution:
Given: A \(\Delta\)
To prove: BD = CE
Proof: In \(\Delta\)
\(\angle\)
and, \(\angle\)
But \(\angle\)
\(\Rightarrow \)
AB = AC [Given]
\(\angle\)
Thus by Angle-Side-Angle criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal.
BD = CE [c.p.c.t]
Question 3:
In the adjoining figure, AD is a median of \(\Delta\)
Solution:
Given: A triangle in which D is the midpoint of BC and BL \(\perp\)
To prove: BL = CM
Proof: In \(\Delta\)
\(\angle\)
\(\angle\)
BD = DC [Given]
Thus by Angle-Angle-Side criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal
So, BL = CM [c.p.c.t]
Question 4:
In \(\Delta\)
Solution:
Given: In a \(\Delta\)
To prove: AB = AC
Proof: In right angled triangle \(\Delta\)
\(\angle\)
Hypt. BD = Hypt. CD [Given]
DL = DM [Given]
Thus, by Right Angle-Hypotenuse_side criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal.
Therefore, \(\angle\)
In \(\Delta\)
\(\angle\)
\(\Rightarrow \)
Question 5:
In \(\Delta\)
Solution:
Given: A \(\Delta\)
To prove: In \(\Delta\)
\(\angle\)
and, \(\angle\)
But, \(\angle\)
So, \(\angle\)
Since base angles are equal, sides are equal
\(\Rightarrow \)
Since OB and OC are the bisectors of angles,
\(\angle\)
\(\angle\)
\(\angle\)
\(\Rightarrow \)
Now, in \(\Delta\)
AB = AC [Given]
\(\angle\)
BO = OC [from(1)]
Thus, by Side-Angle-Side criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal
Therefore, \(\angle\)
i.e. , AO bisects \(\angle\)
Question 6:
In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that
(i) PT = PS
(ii) \(\angle\)
Solution:
Given: PQR is an equilateral triangle and QRST is a square.
To prove: PT = PS
and \(\angle\)
Proof: Since \(\Delta\)
\(\angle\)
Since QRTS is a square,
\(\angle\)
In \(\Delta\)
\(\angle\)
= 60o + 90o = 150o
In \(\Delta\)
\(\angle\)
=60o + 90o = 150o ——- (1)
\(\Rightarrow \)
Thus, in \(\Delta\)
PQ = RS [sides of equilateral triangle PQR]
\(\angle\)
QT = RS [sides of square QRST]
Thus, by Side-Angle-Side criterion of congruence, we have
therefore, \(\Delta\)
The corresponding parts of the congruent triangles are equal.
Therefore, PT = PS [c.p.c.t]
Now in \(\Delta\)
PR = RS
\(\Rightarrow \)
But \(\angle\)
So, by angle sum property in \(\Delta\)
\(\angle\)
\(\Rightarrow \)
\(\Rightarrow \)
\(\Rightarrow \)
Question 7:
In the given figure, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD = BF.
Solution:
Given: ABC is a triangle right angled at B. ACFG is a square and BCDE is a square.
To prove: AD = EF
Proof: Since BCDE is a square,
\(\angle\)
In \(\Delta\)
\(\angle\)
= \(\angle\)
In \(\Delta\)
\(\angle\)
Since ACFG is a square,
\(\angle\)
Thus, we have
\(\angle\)
From (2) and (3), we have
\(\angle\)
Thus in \(\Delta\)
AC = CF [sides of a square]
\(\angle\)
CD = BC [sides of a square]
Thus, by Side-Angle-Side criterion of congruence, we have
Therefore, \(\Delta\)
The corresponding parts of congruent triangles are equal.
So, AD = BF (c.p.c.t)
Question 8:
Prove that the median from the vertex of an isosceles triangle is the bisector of the vertical angle.
Solution:
Given: ABC is a isosceles triangle in which AB = AC and AD is the median through A.
To prove: \(\angle\)
Proof: In \(\Delta\)
AB = AC [Given]
BD = DC [Given]
AD = AD [Common]
Thus by Side-Side-Side criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal.
Therefore, \(\angle\)
Question 9:
In the given figure, ABCD is quadrilateral in which AB || DC and P is the mid-point of BC. On producing, AO and DC meet at Q. Prove that
(i) AB = CQ
(ii) DQ = DC + AB
Solution:
Given ABCD is a quadrilateral in which AB || DC
To prove:
i) AB = CQ
ii) DQ = DC + AB
Proof: In \(\Delta\)
\(\angle\)
\(\angle\)
BP = PC [Given]
Thus by Angle-Angle-Side criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal
Therefore, AB = CQ ——– (1)
Now, DQ = DC + CQ
= DC + AB [from (1)]
Question 10:
In the given figure, OA = OB and OP = OQ. Prove that
(i) PX = QX
(ii) AX = BX
Solution:
Given: OA = OB and OP = OQ
To prove:
(i) PX = QX
(ii) AX = BX
Proof: In \(\Delta\)
OA = OB [Given]
\(\angle\)
OQ = OP [Given]
Thus by Side-Angle-Side criterion of congruence, we have
\(\Delta\)
The corresponding parts of the congruent triangles are equal.
Therefore, \(\angle\)
Thus, in \(\Delta\)
BQ = OB â OQ
and, PA = OA â OP
But, OP = OQ
and OA =OB [Given]
Therefore, we have, BQ = PA ———- (2)
Now consider triangles \(\Delta\)
\(\angle\)
\(\angle\)
BQ = PA [from (2)]
Thus by Angle-Angle-Side criterion of congruence, we have,
Therefore, \(\Delta\)
PX = QX [c.p.c.t]
AX = BX [c.p.c.t]’