Triangles — NCERT Solutions, AI Tutor & Practice

Given: $AC = AD$ and $AB$ bisects $\angle A$, so $\angle CAB = \angle DAB$.

To prove: $\triangle ABC \cong \triangle ABD$.

Proof:

In $\triangle ABC$ and $\triangle ABD$:

1. $AC = AD$ (given)
2. $\angle CAB = \angle DAB$ (since $AB$ bisects $\angle A$)
3. $AB = AB$ (common side)

Therefore, by the SAS congruence rule, $\triangle ABC \cong \triangle ABD.$

Since corresponding parts of congruent triangles are equal (CPCT), $BC = BD.$

Given: $AB = PQ$, $BC = QR$, $CA = RP$.

Since all three pairs of corresponding sides are equal, by the SSS congruence rule, $\triangle ABC \cong \triangle PQR.$

By CPCT, the corresponding angles are equal: $\angle A = \angle P,\quad \angle B = \angle Q,\quad \angle C = \angle R.$

Note: the angle equal to $\angle A$ is the one opposite the side $BC$ (namely $\angle P$, opposite $QR$), so matching must follow the vertex order carefully.

Given: $AB$ and $CD$ bisect each other at $O$, so $AO = BO$ and $CO = DO$.

To prove: $AC = BD$ and $AC \parallel BD$.

Proof:

In $\triangle AOC$ and $\triangle BOD$:

1. $AO = BO$ (O bisects $AB$)
2. $\angle AOC = \angle BOD$ (vertically opposite angles)
3. $CO = DO$ (O bisects $CD$)

By the SAS congruence rule, $\triangle AOC \cong \triangle BOD.$

By CPCT, $AC = BD$.

Also by CPCT, $\angle OAC = \angle OBD$. These are alternate interior angles for lines $AC$ and $BD$ with transversal $AB$, so $AC \parallel BD.$

Given: $AD$ bisects $\angle A$ (so $\angle BAD = \angle CAD$) and $AD \perp BC$ (so $\angle ADB = \angle ADC = 90^\circ$).

To prove: $AB = AC$.

Proof:

In $\triangle ABD$ and $\triangle ACD$:

1. $\angle BAD = \angle CAD$ (given)
2. $AD = AD$ (common side)
3. $\angle ADB = \angle ADC = 90^\circ$ (given)

By the ASA congruence rule, $\triangle ABD \cong \triangle ACD.$

By CPCT, $AB = AC,$ so $\triangle ABC$ is isosceles.

Given: $\angle B = \angle C$ and $\angle BAD = \angle CAD$.

To prove: $BD = CD$.

Proof:

In $\triangle ABD$ and $\triangle ACD$:

1. $\angle B = \angle C$ (given)
2. $\angle BAD = \angle CAD$ (given)
3. $AD = AD$ (common side)

Here two angles and a non-included side are equal, so by the AAS congruence rule, $\triangle ABD \cong \triangle ACD.$

By CPCT, $BD = CD.$

Given: $l$ bisects $\angle A$, so $\angle PAB = \angle QAB$. Also $BP \perp AP$ and $BQ \perp AQ$, so $\angle APB = \angle AQB = 90^\circ$.

To prove: $BP = BQ$.

Proof:

In $\triangle APB$ and $\triangle AQB$:

1. $\angle APB = \angle AQB = 90^\circ$ (given)
2. $\angle PAB = \angle QAB$ (l bisects $\angle A$)
3. $AB = AB$ (common side)

By the AAS congruence rule, $\triangle APB \cong \triangle AQB.$

By CPCT, $BP = BQ.$

Hence $B$ is equidistant from both arms of the angle.

Given: $AB = AC$, and $BO$, $CO$ bisect $\angle B$ and $\angle C$ respectively.

To prove: $OB = OC$.

Proof:

Since $AB = AC$, the angles opposite them are equal (isosceles triangle property): $\angle ABC = \angle ACB.$

Taking halves of equal angles: $\tfrac12\angle ABC = \tfrac12\angle ACB \implies \angle OBC = \angle OCB.$

In $\triangle OBC$, two base angles are equal, so the sides opposite them are equal: $OB = OC.$

Given: $BE \perp AC$, $CF \perp AB$, and $BE = CF$.

To prove: $AB = AC$.

Proof:

In $\triangle BEC$ and $\triangle CFB$:

1. $\angle BEC = \angle CFB = 90^\circ$ (altitudes)
2. $BC = CB$ (common hypotenuse)
3. $BE = CF$ (given)

By the RHS congruence rule, $\triangle BEC \cong \triangle CFB.$

By CPCT, $\angle BCE = \angle CBF,\quad\text{i.e.}\quad \angle ACB = \angle ABC.$

Since the base angles of $\triangle ABC$ are equal, the sides opposite them are equal: $AB = AC.$

Hence $\triangle ABC$ is isosceles.

Given: $AB = AC$ and $AD = AB$, so $AD = AC$. $D$ lies on ray $BA$ produced.

To prove: $\angle BCD = 90^\circ$.

Proof:

In $\triangle ABC$, since $AB = AC$, $\angle ACB = \angle ABC = x \ (\text{say}).$

In $\triangle ACD$, since $AC = AD$, $\angle ACD = \angle ADC = y \ (\text{say}).$

Now $\angle BCD = \angle ACB + \angle ACD = x + y$.

In $\triangle BCD$, the angle sum gives: $\angle DBC + \angle BCD + \angle BDC = 180^\circ.$

Since $\angle DBC = \angle ABC = x$ and $\angle BDC = \angle ADC = y$: $x + (x + y) + y = 180^\circ \implies 2(x + y) = 180^\circ.$

Therefore $x + y = 90^\circ \implies \angle BCD = 90^\circ.$

Given: $\angle A = 40^\circ$, $\angle B = 60^\circ$.

By the angle sum property: $\angle C = 180^\circ - 40^\circ - 60^\circ = 80^\circ.$

So the angles in increasing order are: $\angle A(40^\circ) < \angle B(60^\circ) < \angle C(80^\circ).$

The side opposite a larger angle is longer. The sides opposite these angles are:

• opposite $\angle A$: $BC$
• opposite $\angle B$: $CA$
• opposite $\angle C$: $AB$

Therefore, in ascending order of length: $BC < CA < AB.$

Given: $PQ = 5$ cm, $QR = 7$ cm, $PR = 6$ cm.

Sides in increasing order: $PQ(5) < PR(6) < QR(7).$

The angle opposite a longer side is larger. The angle opposite each side is:

• opposite $PQ$: $\angle R$
• opposite $PR$: $\angle Q$
• opposite $QR$: $\angle P$

Therefore, in ascending order: $\angle R < \angle Q < \angle P.$

Triangle inequality property: In any triangle, the sum of the lengths of any two sides is greater than the length of the third side. This is because the straight segment between two vertices is the shortest path, so going via a third vertex is always longer.

Check for sides $3$, $4$, $8$:

Test the three inequalities:

1. $3 + 4 = 7$, and $7 < 8$ — fails (must be greater than $8$).
2. $4 + 8 = 12 > 3$ — holds.
3. $3 + 8 = 11 > 4$ — holds.

Since the first inequality fails ($3 + 4 \not> 8$), such a triangle cannot exist.