Extra Practice Questions
Triangles (R.D. Sharma Level – Class 9)
1. In quadrilateral ACBD, AC = AD and BC = BD. Prove that AB is the perpendicular bisector of CD.
Proof:
First, in ΔACB and ΔADB:
• AC = AD (Given)
• BC = BD (Given)
• AB = AB (Common)
By SSS congruence, ΔACB ≅ ΔADB. So, ∠CAB = ∠DAB (by CPCT).
Now, let AB and CD intersect at O. In ΔAOC and ΔAOD:
• AC = AD (Given)
• ∠CAO = ∠DAO (Proved above)
• AO = AO (Common)
By SAS congruence, ΔAOC ≅ ΔAOD.
By CPCT, OC = OD (so AB bisects CD) and ∠AOC = ∠AOD.
Since ∠AOC + ∠AOD = 180° (Linear Pair), we have 2∠AOC = 180° => ∠AOC = 90°.
Thus, AB is perpendicular to CD.
Since AB is perpendicular to CD and bisects it, AB is the perpendicular bisector of CD. (Proved)
2. In ΔABC, D is a point on side AC such that AB = AD. Prove that BC > CD.
Proof:
In ΔABD, since AB = AD (Given), the angles opposite to these sides are equal.
Therefore, ∠ABD = ∠ADB.
Now, consider ∠ADB. It is an exterior angle to ΔBDC.
An exterior angle of a triangle is greater than either of its interior opposite angles.
So, ∠ADB > ∠BCD (or ∠C).
From the first step, we can substitute ∠ABD for ∠ADB:
∠ABD > ∠C.
Since ∠ABC = ∠ABD + ∠CBD, it is clear that ∠ABC > ∠ABD.
Combining these inequalities, we get ∠ABC > ∠ABD > ∠C.
This means ∠ABC > ∠C.
In the larger triangle ΔABC, the side opposite the greater angle is longer.
Since ∠ABC > ∠ACB, the side AC > AB.
As AC = AD + DC and AB = AD (given), we have:
AD + DC > AB => AD + DC > AD => DC > 0. (This confirms geometry but doesn’t prove the result).
A much simpler approach:
In ΔABC, by the Triangle Inequality Theorem, the sum of two sides is greater than the third side.
AB + AC > BC.
Let’s apply it to ΔABD: AB + BD > AD. Since AB = AD, this doesn’t help.
Let’s use the exterior angle theorem again, but focus on ΔABC.
Wait, let’s reconsider the very first approach: In ΔABD, AB=AD implies ∠ABD = ∠ADB.
In ΔBDC, the exterior angle at D would be ∠BDA. No.
The simplest proof is: In ΔABC, side AC = AD + DC. Since AB=AD, we can write AC = AB + DC.
By Triangle Inequality in ΔABC: AB + BC > AC.
Substitute AC: AB + BC > AB + DC.
Subtracting AB from both sides: BC > DC.
(Proved)
3. In ΔPQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is the longest? Give reason for your answer.
First, find the third angle, ∠Q.
∠P + ∠Q + ∠R = 180° => 70° + ∠Q + 30° = 180° => ∠Q = 80°.
The angles are ∠Q=80°, ∠P=70°, ∠R=30°.
The side opposite the largest angle is the longest side. The largest angle is ∠Q.
The side opposite to ∠Q is PR.
Answer: The side PR is the longest because it is opposite the largest angle (80°).
4. In ΔABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
First, find ∠C: ∠C = 180° – (40° + 60°) = 80°.
The angles are ∠C=80°, ∠B=60°, ∠A=40°.
The longest side is opposite the largest angle (∠C). So, the longest side is AB.
The shortest side is opposite the smallest angle (∠A). So, the shortest side is BC.
5. In a right-angled triangle, prove that the hypotenuse is the longest side.
Given: A right-angled triangle ΔABC, with ∠B = 90°.
To Prove: Hypotenuse AC is the longest side.
Proof:
In ΔABC, ∠A + ∠B + ∠C = 180°.
∠A + 90° + ∠C = 180° => ∠A + ∠C = 90°.
Since ∠A and ∠C are positive, both must be less than 90°.
So, ∠B is the largest angle of the triangle.
The side opposite the largest angle is the longest. The side opposite ∠B is the hypotenuse AC.
Therefore, the hypotenuse is the longest side. (Proved)
6. In quadrilateral ABCD, the diagonal AC bisects the angles ∠A and ∠C. Prove that AB = AD and CB = CD.
Proof:
Consider ΔABC and ΔADC.
• ∠BAC = ∠DAC (AC bisects ∠A)
• AC = AC (Common side)
• ∠BCA = ∠DCA (AC bisects ∠C)
By ASA congruence rule, ΔABC ≅ ΔADC.
Since the triangles are congruent, their corresponding parts are equal.
Therefore, AB = AD and CB = CD (by CPCT).
(Proved)
7. In quadrilateral ABCD, AD = BC and BD = CA. Prove that ∠ADB = ∠BCA.
Proof:
Consider ΔADB and ΔBCA.
• AD = BC (Given)
• BD = CA (Given)
• AB = BA (Common side)
By SSS congruence rule, ΔADB ≅ ΔBCA.
Since the triangles are congruent, their corresponding angles are equal.
Therefore, ∠ADB = ∠BCA (by CPCT).
(Proved)
8. In parallelogram ABCD, E is the mid-point of BC. DE and AB when produced meet at F. Prove that AF = 2AB.
Proof:
Consider ΔDCE and ΔFBE.
• ∠DEC = ∠FEB (Vertically opposite angles)
• CE = BE (E is the mid-point of BC)
• ∠DCE = ∠FBE (Alternate interior angles, since DC || AF)
By ASA congruence rule, ΔDCE ≅ ΔFBE.
By CPCT, DC = FB.
Since ABCD is a parallelogram, DC = AB.
Therefore, FB = AB.
Now, AF = AB + FB = AB + AB = 2AB.
(Proved)
9. Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD.
Proof:
We use the property that the sum of two sides of a triangle is greater than the third side.
In ΔABC, AB + BC > AC —(1)
In ΔADC, AD + DC > AC —(2)
In ΔABD, AB + AD > BD —(3)
In ΔBCD, BC + CD > BD —(4)
Adding all four inequalities:
(AB+BC) + (AD+DC) + (AB+AD) + (BC+CD) > 2AC + 2BD
2AB + 2BC + 2CD + 2AD > 2(AC + BD)
2(AB + BC + CD + DA) > 2(AC + BD)
AB + BC + CD + DA > AC + BD.
(Proved)
10. In ΔABC, D is any point on side BC. Show that AB + BC + CA > 2AD.
Proof:
Apply the triangle inequality to the two triangles formed by the line segment AD.
In ΔABD, the sum of two sides is greater than the third side.
AB + BD > AD —(1)
In ΔACD, the sum of two sides is greater than the third side.
AC + CD > AD —(2)
Adding inequalities (1) and (2):
(AB + BD) + (AC + CD) > AD + AD
AB + AC + (BD + CD) > 2AD.
Since BD + CD = BC, we have:
AB + AC + BC > 2AD.
(Proved)
