Triangles

Given: In quadrilateral ACBD, AC = AD and AB bisects ∠A, i.e., ∠CAB = ∠DAB.

To prove: △ABC ≅ △ABD

Proof:

Consider △ABC and △ABD.

$$AC = AD \quad \text{(Given)}$$

$$\angle CAB = \angle DAB \quad \text{(AB bisects } \angle A\text{)}$$

$$AB = AB \quad \text{(Common side)}$$

Therefore, by SAS congruence rule:
$$\Delta ABC \cong \Delta ABD$$

About BC and BD:
Since △ABC ≅ △ABD, by CPCT:
$$BC = BD$$
So BC and BD are equal, i.e., B is equidistant from C and D.

Given: In quadrilateral ABCD, AD = BC and ∠DAB = ∠CBA.

(i) To prove: △ABD ≅ △BAC

Consider △ABD and △BAC.

$$AD = BC \quad \text{(Given)}$$

$$\angle DAB = \angle CBA \quad \text{(Given)}$$

$$AB = BA \quad \text{(Common side)}$$

Therefore, by SAS congruence rule:
$$\Delta ABD \cong \Delta BAC$$

(ii) To prove: BD = AC

Since △ABD ≅ △BAC (proved above), by CPCT:
$$BD = AC$$

(iii) To prove: ∠ABD = ∠BAC

Since △ABD ≅ △BAC, by CPCT:
$$\angle ABD = \angle BAC$$

Given: AD ⊥ AB, BC ⊥ AB, and AD = BC.

To prove: CD bisects AB, i.e., the point of intersection O of CD and AB is the mid-point of AB (OA = OB).

Proof:

Consider △AOD and △BOC.

$$\angle AOD = \angle BOC \quad \text{(Vertically opposite angles)}$$

$$\angle DAO = \angle CBO = 90^\circ \quad \text{(AD } \perp \text{ AB and BC } \perp \text{ AB)}$$

$$AD = BC \quad \text{(Given)}$$

Therefore, by AAS congruence rule:
$$\Delta AOD \cong \Delta BOC$$

By CPCT:
$$OA = OB$$

Hence, O is the mid-point of AB, i.e., CD bisects AB.

Given: $l \parallel m$ and $p \parallel q$. ABCD is a parallelogram formed by these lines, with AC as the diagonal.

To prove: △ABC ≅ △CDA

Proof:

Consider △ABC and △CDA.

Since $p \parallel q$ and AC is a transversal:
$$\angle BAC = \angle DCA \quad \text{(Alternate interior angles)}$$

Since $l \parallel m$ and AC is a transversal:
$$\angle BCA = \angle DAC \quad \text{(Alternate interior angles)}$$

$$AC = CA \quad \text{(Common side)}$$

Therefore, by ASA congruence rule:
$$\Delta ABC \cong \Delta CDA$$

Given: Line $l$ bisects ∠A, so ∠PAB = ∠QAB. BP ⊥ AP and BQ ⊥ AQ, so ∠APB = ∠AQB = 90°.

(i) To prove: △APB ≅ △AQB

Consider △APB and △AQB.

$$\angle APB = \angle AQB = 90^\circ \quad \text{(BP and BQ are perpendiculars)}$$

$$\angle PAB = \angle QAB \quad \text{(}l\text{ bisects }\angle A\text{)}$$

$$AB = AB \quad \text{(Common hypotenuse)}$$

Therefore, by AAS congruence rule:
$$\Delta APB \cong \Delta AQB$$

(ii) To prove: BP = BQ

Since △APB ≅ △AQB, by CPCT:
$$BP = BQ$$

Hence, B is equidistant from the arms of ∠A.

Given: AC = AE, AB = AD, and ∠BAD = ∠EAC.

To prove: BC = DE

Proof:

We have:
$$\angle BAD = \angle EAC \quad \text{(Given)}$$

Adding ∠DAC to both sides:
$$\angle BAD + \angle DAC = \angle EAC + \angle DAC$$

$$\angle BAC = \angle DAE$$

Now consider △ABC and △ADE.

$$AB = AD \quad \text{(Given)}$$

$$\angle BAC = \angle DAE \quad \text{(Proved above)}$$

$$AC = AE \quad \text{(Given)}$$

Therefore, by SAS congruence rule:
$$\Delta ABC \cong \Delta ADE$$

By CPCT:
$$BC = DE$$

Given: P is the mid-point of AB, so AP = BP. ∠BAD = ∠ABE and ∠EPA = ∠DPB.

(i) To prove: △DAP ≅ △EBP

We have:
$$\angle EPA = \angle DPB \quad \text{(Given)}$$

Adding ∠EPD to both sides:
$$\angle EPA + \angle EPD = \angle DPB + \angle EPD$$

$$\angle DPA = \angle EPB$$

Now consider △DAP and △EBP.

$$\angle DAP = \angle EBP \quad \text{(Given: }\angle BAD = \angle ABE\text{)}$$

$$AP = BP \quad \text{(P is mid-point of AB)}$$

$$\angle DPA = \angle EPB \quad \text{(Proved above)}$$

Therefore, by ASA congruence rule:
$$\Delta DAP \cong \Delta EBP$$

(ii) To prove: AD = BE

Since △DAP ≅ △EBP, by CPCT:
$$AD = BE$$

Given: △ABC is right-angled at C. M is the mid-point of AB, so AM = BM. CM is produced to D such that DM = CM.

(i) To prove: △AMC ≅ △BMD

Consider △AMC and △BMD.

$$AM = BM \quad \text{(M is mid-point of AB)}$$

$$\angle AMC = \angle BMD \quad \text{(Vertically opposite angles)}$$

$$CM = DM \quad \text{(Given)}$$

Therefore, by SAS congruence rule:
$$\Delta AMC \cong \Delta BMD$$

(ii) To prove: ∠DBC is a right angle

From (i), by CPCT:
$$\angle MAC = \angle MBD$$

These are alternate interior angles for lines AC and DB with transversal BC.

Therefore, $AC \parallel DB$, which means:
$$\angle DBC + \angle ACB = 180^\circ \quad \text{(Co-interior angles)}$$

Since ∠ACB = 90° (given):
$$\angle DBC = 180^\circ - 90^\circ = 90^\circ$$

Hence, ∠DBC is a right angle.

(iii) To prove: △DBC ≅ △ACB

Consider △DBC and △ACB.

$$DB = AC \quad \text{(CPCT from part (i))}$$

$$\angle DBC = \angle ACB = 90^\circ \quad \text{(Proved above and given)}$$

$$BC = CB \quad \text{(Common side)}$$

Therefore, by SAS congruence rule:
$$\Delta DBC \cong \Delta ACB$$

(iv) To prove: CM = ½ AB

From (iii), by CPCT:
$$DC = AB$$

But $DC = DM + CM = CM + CM = 2\,CM$ (since DM = CM).

Therefore:
$$2\,CM = AB$$

$$\boxed{CM = \dfrac{1}{2}\,AB}$$

Given: △ABC is isosceles with AB = AC. Bisectors of ∠B and ∠C meet at O.

Since AB = AC, the angles opposite to equal sides are equal:
$$\angle ABC = \angle ACB$$

Therefore:
$$\frac{1}{2}\angle ABC = \frac{1}{2}\angle ACB$$

$$\angle OBC = \angle OCB \quad \text{(BO and CO are bisectors)}$$

(i) To prove: OB = OC

In △OBC:
$$\angle OBC = \angle OCB \quad \text{(Proved above)}$$

Sides opposite to equal angles are equal:
$$OB = OC$$

(ii) To prove: AO bisects ∠A

Consider △ABO and △ACO.

$$AB = AC \quad \text{(Given)}$$

$$OB = OC \quad \text{(Proved in part (i))}$$

$$AO = AO \quad \text{(Common side)}$$

Therefore, by SSS congruence rule:
$$\Delta ABO \cong \Delta ACO$$

By CPCT:
$$\angle BAO = \angle CAO$$

Hence, AO bisects ∠A.

Given: AD is the perpendicular bisector of BC, so BD = DC and ∠ADB = ∠ADC = 90°.

To prove: AB = AC

Proof:

Consider △ADB and △ADC.

$$BD = DC \quad \text{(AD bisects BC)}$$

$$\angle ADB = \angle ADC = 90^\circ \quad \text{(AD } \perp \text{ BC)}$$

$$AD = AD \quad \text{(Common side)}$$

Therefore, by SAS congruence rule:
$$\Delta ADB \cong \Delta ADC$$

By CPCT:
$$AB = AC$$

Hence, △ABC is an isosceles triangle.

Given: △ABC is isosceles with AB = AC. BE ⊥ AC and CF ⊥ AB.

To prove: BE = CF

Proof:

Consider △ABE and △ACF.

$$\angle AEB = \angle AFC = 90^\circ \quad \text{(BE and CF are altitudes)}$$

$$\angle A = \angle A \quad \text{(Common angle)}$$

$$AB = AC \quad \text{(Given)}$$

Therefore, by AAS congruence rule:
$$\Delta ABE \cong \Delta ACF$$

By CPCT:
$$BE = CF$$

Hence, the altitudes are equal.

Given: BE ⊥ AC and CF ⊥ AB, and BE = CF.

(i) To prove: △ABE ≅ △ACF

Consider △ABE and △ACF.

$$\angle AEB = \angle AFC = 90^\circ \quad \text{(BE and CF are altitudes)}$$

$$\angle A = \angle A \quad \text{(Common angle)}$$

$$BE = CF \quad \text{(Given)}$$

Therefore, by AAS congruence rule:
$$\Delta ABE \cong \Delta ACF$$

(ii) To prove: AB = AC

Since △ABE ≅ △ACF, by CPCT:
$$AB = AC$$

Hence, △ABC is an isosceles triangle.

Given: △ABC is isosceles with AB = AC. △DBC is isosceles with DB = DC.

To prove: ∠ABD = ∠ACD

Proof:

Consider △ABD and △ACD.

$$AB = AC \quad \text{(△ABC is isosceles)}$$

$$DB = DC \quad \text{(△DBC is isosceles)}$$

$$AD = AD \quad \text{(Common side)}$$

Therefore, by SSS congruence rule:
$$\Delta ABD \cong \Delta ACD$$

By CPCT:
$$\angle ABD = \angle ACD$$

Given: △ABC is isosceles with AB = AC. BA is produced to D such that AD = AB.

To prove: ∠BCD = 90°

Proof:

Since AB = AC, angles opposite to equal sides are equal:
$$\angle ABC = \angle ACB \quad \ldots (1)$$

Now, AD = AB = AC, so in △ACD:
$$AD = AC$$

Angles opposite to equal sides are equal:
$$\angle ACD = \angle ADC \quad \ldots (2)$$

In △BCD, the sum of angles is 180°:
$$\angle DBC + \angle BDC + \angle BCD = 180^\circ$$

Note that ∠DBC = ∠ABC (same angle) and ∠BDC = ∠ADC.

So:
$$\angle ABC + \angle ADC + \angle BCD = 180^\circ$$

Using (1) and (2):
$$\angle ACB + \angle ACD + \angle BCD = 180^\circ$$

$$\angle BCD + \angle BCD = 180^\circ$$

$$2\angle BCD = 180^\circ$$

$$\angle BCD = 90^\circ$$

Hence, ∠BCD is a right angle.

Given: In △ABC, ∠A = 90° and AB = AC.

To find: ∠B and ∠C

Since AB = AC, the angles opposite to equal sides are equal:
$$\angle ABC = \angle ACB$$

Let ∠ABC = ∠ACB = $x$.

By the angle sum property of a triangle:
$$\angle A + \angle B + \angle C = 180^\circ$$

$$90^\circ + x + x = 180^\circ$$

$$2x = 90^\circ$$

$$x = 45^\circ$$

Therefore:
$$\boxed{\angle B = \angle C = 45^\circ}$$

Given: △ABC is equilateral, so AB = BC = CA.

To prove: ∠A = ∠B = ∠C = 60°

Proof:

Since AB = BC, angles opposite to equal sides are equal:
$$\angle BAC = \angle BCA \quad \ldots (1)$$

Since BC = CA, angles opposite to equal sides are equal:
$$\angle BAC = \angle ABC \quad \ldots (2)$$

From (1) and (2):
$$\angle BAC = \angle ABC = \angle BCA$$

Let each angle = $x$.

By angle sum property:
$$x + x + x = 180^\circ$$

$$3x = 180^\circ$$

$$x = 60^\circ$$

Therefore, each angle of an equilateral triangle is 60°.

Given: △ABC is isosceles with AB = AC. △DBC is isosceles with DB = DC. AD extended meets BC at P.

(i) To prove: △ABD ≅ △ACD

Consider △ABD and △ACD.

$$AB = AC \quad \text{(△ABC is isosceles)}$$

$$DB = DC \quad \text{(△DBC is isosceles)}$$

$$AD = AD \quad \text{(Common side)}$$

Therefore, by SSS congruence rule:
$$\Delta ABD \cong \Delta ACD$$

By CPCT: $\angle BAD = \angle CAD$, i.e., AD bisects ∠A. Also $\angle BDA = \angle CDA$, i.e., AD bisects ∠D.

(ii) To prove: △ABP ≅ △ACP

Consider △ABP and △ACP.

$$AB = AC \quad \text{(Given)}$$

$$\angle BAP = \angle CAP \quad \text{(CPCT from part (i), }\angle BAD = \angle CAD\text{)}$$

$$AP = AP \quad \text{(Common side)}$$

Therefore, by SAS congruence rule:
$$\Delta ABP \cong \Delta ACP$$

(iii) AP bisects ∠A as well as ∠D

- From part (i): $\angle BAD = \angle CAD$, so AP (i.e., AD) bisects ∠A.
- From part (i): $\angle BDA = \angle CDA$, so AP bisects ∠D.

(iv) AP is the perpendicular bisector of BC

From part (ii), by CPCT:
$$BP = CP \quad \text{(AP bisects BC)}$$

Also from part (ii), by CPCT:
$$\angle APB = \angle APC$$

Since ∠APB + ∠APC = 180° (linear pair):
$$2\angle APB = 180^\circ \implies \angle APB = 90^\circ$$

Therefore, AP ⊥ BC and AP bisects BC.

Hence, AP is the perpendicular bisector of BC.

Given: △ABC is isosceles with AB = AC. AD ⊥ BC (AD is altitude), so ∠ADB = ∠ADC = 90°.

(i) To prove: AD bisects BC, i.e., BD = DC

Consider △ADB and △ADC.

$$AB = AC \quad \text{(Given)}$$

$$\angle ADB = \angle ADC = 90^\circ \quad \text{(AD is altitude)}$$

$$AD = AD \quad \text{(Common side)}$$

Therefore, by RHS congruence rule:
$$\Delta ADB \cong \Delta ADC$$

By CPCT:
$$BD = DC$$

Hence, AD bisects BC.

(ii) To prove: AD bisects ∠A

From △ADB ≅ △ADC (proved above), by CPCT:
$$\angle BAD = \angle CAD$$

Hence, AD bisects ∠A.

Given: AB = PQ, BC = QR, AM is median of △ABC (so BM = ½BC), PN is median of △PQR (so QN = ½QR), and AM = PN.

(i) To prove: △ABM ≅ △PQN

Since AM is the median of △ABC:
$$BM = \frac{1}{2}BC$$

Since PN is the median of △PQR:
$$QN = \frac{1}{2}QR$$

Given BC = QR, therefore:
$$BM = QN$$

Now consider △ABM and △PQN.

$$AB = PQ \quad \text{(Given)}$$

$$BM = QN \quad \text{(Proved above)}$$

$$AM = PN \quad \text{(Given)}$$

Therefore, by SSS congruence rule:
$$\Delta ABM \cong \Delta PQN$$

(ii) To prove: △ABC ≅ △PQR

From part (i), by CPCT:
$$\angle ABM = \angle PQN$$

i.e., $\angle ABC = \angle PQR$

Now consider △ABC and △PQR.

$$AB = PQ \quad \text{(Given)}$$

$$\angle ABC = \angle PQR \quad \text{(Proved above)}$$

$$BC = QR \quad \text{(Given)}$$

Therefore, by SAS congruence rule:
$$\Delta ABC \cong \Delta PQR$$

Given: In △ABC, BE ⊥ AC and CF ⊥ AB, and BE = CF.

To prove: △ABC is isosceles, i.e., AB = AC.

Proof:

Consider △BCF and △CBE.

$$\angle BFC = \angle CEB = 90^\circ \quad \text{(CF and BE are altitudes)}$$

$$BC = CB \quad \text{(Common hypotenuse)}$$

$$CF = BE \quad \text{(Given)}$$

Therefore, by RHS congruence rule:
$$\Delta BCF \cong \Delta CBE$$

By CPCT:
$$\angle FBC = \angle ECB$$

i.e., $\angle ABC = \angle ACB$

Sides opposite to equal angles are equal:
$$AC = AB$$

Hence, △ABC is isosceles.

Given: △ABC is isosceles with AB = AC. AP ⊥ BC, so ∠APB = ∠APC = 90°.

To prove: ∠B = ∠C

Proof:

Consider △APB and △APC.

$$AB = AC \quad \text{(Given)}$$

$$\angle APB = \angle APC = 90^\circ \quad \text{(AP } \perp \text{ BC)}$$

$$AP = AP \quad \text{(Common side)}$$

Therefore, by RHS congruence rule:
$$\Delta APB \cong \Delta APC$$

By CPCT:
$$\angle ABP = \angle ACP$$

i.e., $\angle B = \angle C$

Hence proved.