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.