Triangles

Given:

As we know, two triangles are congruent, If the side (S) and angles (A) of one triangle is equal to another angle.

And criterion for congruence of triangle are SAS, ASA, SSS, and RHS.

SSA is not the criterion for congruency of a triangle.

Hence, option C is correct.

Given: AB = QR, BC = PR and CA = PQ

As AB = QR, BC = PR and CA = PQ

⇒ A corresponds to Q, B corresponds to R, C corresponds to P.

i.e.

Hence, ΔCBA ≅Δ PRQ.

Hence, option B is correct.

Given: Δ ABC, AB = AC and ∠B = 50°.

As AB = AC

So it is an isosceles triangle.

So ∠B = ∠C

∠B = 50° (given)

⇒ ∠C = 50°

Hence, option B is correct.

Given: Δ ABC, BC = AB and ∠B = 80°

As BC = AB

So it is an isosceles triangle.

let ∠C = ∠A = x

∠B = 80° (given)

As we know, ∠A + ∠B + ∠C = 180°

⇒ x + 80° + x = 180°

⇒ 2x = 180° - 80°

⇒ 2x = 100°

⇒ x = 50°

So, ∠C = ∠A = 50°

Hence, option C is correct.

Given: ΔPQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm

Because, ∠R = ∠P

So it is an isosceles triangle.

Hence, PQ = QR

⇒ PQ = 4cm

Hence, option A is correct.

Given: ΔABC such that AD bisects ∠BAC

As AD bisects ∠BAC hence, ∠BAD = ∠CAD

In ΔACD, ∠ADB is exterior angle.

Hence, ∠ADB > ∠DAC (∵ exterior angle is greater than interior angle)

⇒ ∠ADB > ∠BAD (∵ ∠BAD = ∠CAD)

⇒ BA > BD (side opposite to greater angle is greater)

Hence, option B is correct.

Given: two sides are given.

Let AB = 5cm and BC = 1.5cm

Third side AC should:

AB-BC < AC < AB + BC

5-1.5 < AC < 5 + 1.5

3.5 < AC < 6.5

Hence, it cannot be 3.4 because it is less than 3.5.

Option D is correct.

Given: In ΔPQR, ∠R >∠Q

⇒ PQ > PR (side opposite to greater angle is greater)

Hence, B is correct.

Given: ΔABC and ΔPQR, AB = AC, ∠C = ∠P and ∠B = ∠Q

AB = AC

⇒ ∠B = ∠C (opposite angles to equal sides are equal)

Hence, ΔABC is an isosceles triangle.

∠C = ∠P and ∠B = ∠Q (given)

⇒ ∠P = ∠Q (∵∠B = ∠C)

⇒ QR = PR (opposite sides to equal angles are equal)

Hence, ΔPQR is an isosceles triangle.

So, the two triangles are isosceles but not congruent.

As AAA is not the criterion for a triangle to be congruent.

Hence, option A is correct.

Given: ΔABC and ΔDEF, AB = FD and ∠A = ∠D

The two triangles will be congruent by SAS axiom if two sides including one angle of one triangle is equal to the another triangle.

Hence, AC = DE (As ∠A is between AB and AC)

Option B is correct.

Given: ΔABC is an isosceles triangle and AB = AC, BD and CE are two medians.

In ΔABD and ΔACE,

AB = AC (given)

We can write it as:

2 AE = 2 AD (as D and E are mid points)

So, AE = AD

∠A = ∠A (common)

Hence, ΔABD ≅ ΔACE (using SAS)

⇒ BD = CE (by CPCT)

Hence proved.

Given: In ΔABC, BD = CE and AD = AE.

In ΔADE,

AD = AE

⇒ ∠ADE = ∠AED …(1) (opposite angles to equal sides are equal)

Now, ∠ADE + ∠ADB = 180° (linear pair)

∠ADB = 180° - ∠ADE ..(2)

Also, ∠AED + ∠AEC = 180° (linear pair)

∠AEC = 180° - ∠AED

∠AEC = 180° - ∠ADE .. (3) (∵∠ADE = ∠AED)

From (2) and (3)

∠ADB = ∠AEC ..(4)

Now, In ΔADB and ΔAEC,

AD = AE (given)

BD = EC (given)

∠ADB = ∠AEC (from (4)

Hence, ΔABD ≅ΔACE (by SAS)

Given: CDE is an equilateral triangle formed on a side CD of a square ABCD.

In ΔADE and ΔBCE,

DE = CE (sides of equilateral triangle)

Now, ∠ADC = ∠BCD = 90° and ∠EDC = ∠ECD = 60°

Hence, ∠ADE = ∠ADC + ∠CDE = 90° + 60° = 150°

And ∠BCE = ∠BCD + ∠ECD = 90° + 60° = 150°

⇒ ∠ADE = ∠BCE

AD = BC (sides of square)

Hence, ΔADE ≅ΔBCE (by SAS)

Given: BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC.

In ΔABC and ΔDEF

BA = DE (given)

BF = EC(given)

∠A = ∠D (both 90°)

BC = BF + FC

EF = EC + FC = BF + FC (∵ EC = BF)

⇒ EF = BC

Hence, ΔABC ≅ΔDEF (by RHS)

Given: in ΔPSR, Q is a point on the side SR such that PQ = PR.

In ΔPRQ,

PR = PQ (given)

⇒ ∠PRQ = ∠PQR (opposite angles to equal sides are equal)

But ∠PQR > ∠PSR (exterior angle of a triangle is greater than each of opposite interior angle)

⇒ ∠PRQ > ∠PSR

⇒ PS > PR (opposite sides to greater angle is greater)

⇒ PS > PQ (as PR = PQ)

Given: S is any point on side QR of a ΔPQR

In ΔPQS,

PQ + QS > PS (sum of two sides is greater than the third side) ...(1)

Similarly, In ΔPRS,

SR + RP > PS (sum of two sides is greater than the third side) ...(2)

Add (1) and (2)

PQ + QS + SR + RP > 2 PS

⇒ PQ + (QS + SR) + RP > 2 PS

⇒ PQ + QR + RP > 2 PS (as, QS + SR = QR)

Hence, proved.

Given: in ΔABC, AB = AC

In ΔABC,

AC = AB

∠ABC = ∠ACB (opposite angles to equal sides are equal)

In ΔABC and ΔDBC,

∠ABC > ∠DBC (since ∠DBC is interior angle of ∠ABC)

⇒ ∠ACB > ∠DBC (∵ ∠ABC = ∠ACB)

⇒ BD > CD (opposite sides to greater angle is greater)

Or CD < BD

Given: L || m and M is mid point of segment AB.

As L || m,

∠BAC = ∠ABD (alternate interior angle)

∠AMC = ∠DMB (vertically opposite angle)

In ΔAMC and ΔBMD

∠BAC = ∠ABD (proved)

∠AMC = ∠DMB (proved)

AM = BM (given)

Hence, ΔAMC ≅ΔBMD (by ASA)

⇒ MC = MD (by CPCT)

Given: Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O.

In ΔABC,

AB = AC (given)

⇒ ∠ACB = ∠ABC (opposite angles to equal sides are equal)

1/2 ∠ACB = 1/2 ∠ABC (divide both sides by 2)

⇒ ∠OCB = ∠OBC …(1) (As OB and OC are bisector of ∠B and ∠C)

Now, ∠MOC = ∠OBC + ∠OCB (as exterior angle is equal to sum of two opposite interior angle)

⇒∠MOC = ∠OBC + ∠OBC (from (1))

⇒ ∠MOC = 2∠OBC

⇒∠MOC = ∠ABC (because OB is bisector of ∠B)

Hence proved.

Given: ΔABC with AB = AC

In ΔABC,

AB = AC (given)

⇒ ∠ACB = ∠ABC (opposite angles to equal sides are equal)

1/2 ∠ACB = 1/2 ∠ABC (divide both sides by 2)

⇒ ∠OCB = ∠OBC …(1) (As OB and OC are bisector of ∠B and ∠C)

Now, in ΔBOC,

∠OBC + ∠OCB + ∠BOC = 180° (angle sum property)

∠OBC + ∠OBC + ∠BOC = 180° (from (1))

⇒ 2 ∠OBC + ∠BOC = 180°

⇒ ∠ABC + ∠BOC = 180° (because OB is bisector of ∠B)

⇒ 180° - ∠DBA + ∠BOC = 180°

⇒ - ∠DBA + ∠BOC = 0

⇒∠DBA = ∠BOC Hence proved.

Given: AD is the bisector of ∠BAC.

As AD is the bisector of ∠BAC.

∠BAD = ∠CAD …(1)

∠ADB > ∠CAD (exterior angle of a triangle is greater than each of opposite interior angle)

Then ∠ADB > ∠BAD (from (1))

⇒ AB > BD (opposite sides to greater angle is greater)

Hence proved.