✅ Proof

Step 1: Given information

Given: ΔABC ∼ ΔFEG
∴ ∠A = ∠F, ∠B = ∠E, ∠C = ∠G
Also: CD bisects ∠ACB ⇒ ∠ACD = ∠DCB = ½∠C
GH bisects ∠EGF ⇒ ∠EGH = ∠HGF = ½∠G

Step 2: Show corresponding angles are equal

In ΔDCA and ΔHGF:
1. ∠A = ∠F (from ΔABC ∼ ΔFEG)
2. ∠ACD = ½∠C = ½∠G = ∠HGF
(Since ∠C = ∠G and both are bisected)

Step 3: Apply AA similarity criterion

In ΔDCA and ΔHGF:
∠A = ∠F
∠ACD = ∠HGF
∴ By AA similarity criterion:
ΔDCA ∼ ΔHGF

✅ Proof

Given:

• ΔFEC ≅ ΔGDB
• ∠1 = ∠2

To Prove:

ΔADE ~ ΔABC

Proof:

Given ∠1 = ∠2

AD = AE   ( side opposite to equal angle are equal ) —-l

Since ΔFEC ≅ ΔGDB, their corresponding parts are equal.

Therefore:

EC = DB —-ll

From l and ll

AD / DB = AE / EC

By converse of B.P.T

Hence, DE ∥ BC

Using Parallel Lines Property:

∠1 = ∠3.   [ By corresponding angles are equal ]

∠2 = ∠4     [ By corresponding angles are equal ]

In ΔADE and ΔABC

• ∠1 = ∠3 [ proved above ]
• ∠2 = ∠4 [ proved above ]

Therefore, by AA Similarity Rule:

ΔADE ~ ΔABC

Hence Proved ✔

✅ Proof

Step 1: Identify similar triangles

In ΔAPB and ΔDPC:
1. ∠APB = ∠DPC (vertically opposite angles)
2. ∠BAP = ∠CDP ( 90 degree each )
3. ∠ABP = ∠DCP (remaining angles)
∴ ΔAPB ∼ ΔDPC (by AA similarity)

Step 2: Use property of similar triangles

From ΔAPB ∼ ΔDPC:
Corresponding sides are proportional:
AP/DP = PB/PC = AB/DC

Step 3: Prove required relation

From AP/DP = PB/PC
Cross multiply: AP × PC = PB × DP
∴ AP × PC = BP × DP

✅ Proof (Basic Proportionality Theorem/Thales Theorem)

Given:

In ΔABC, DE || BC, intersecting AB at D and AC at E.
To prove: AD/DB = AE/EC

Proof:

1. Join BE and CD.
2. Draw EL ⊥ AB and DM ⊥ AC.
3. Area of ΔADE = ½ × AD × EL
Area of ΔBDE = ½ × DB × EL
∴ Area(ΔADE)/Area(ΔBDE) = AD/DB … (i)

4. Similarly, Area(ΔADE) = ½ × AE × DM
Area(ΔCDE) = ½ × EC × DM
∴ Area(ΔADE)/Area(ΔCDE) = AE/EC … (ii)

5. But ΔBDE and ΔCDE are on same base DE and between same parallels DE and BC.
∴ Area(ΔBDE) = Area(ΔCDE)
⇒ Area(ΔADE)/Area(ΔBDE) = Area(ΔADE)/Area(ΔCDE)

6. From (i) and (ii):
AD/DB = AE/EC

Hence proved.

✅ Proof

Step 1: Given information

Given: ΔABC ∼ ΔPQR
∴ AB/PQ = BC/QR = AC/PR … (i)
Also: AD is median of ΔABC ⇒ BD = DC = ½BC
PM is median of ΔPQR ⇒ QM = MR = ½QR

Step 2: Show ΔABD ∼ ΔPQM

In ΔABD and ΔPQM:
1. ∠B = ∠Q (from ΔABC ∼ ΔPQR)
2. AB/PQ = BC/QR (from i)
But BC/QR = 2BD/2QM = BD/QM
∴ AB/PQ = BD/QM

Step 3: Apply SAS similarity

In ΔABD and ΔPQM:
∠B = ∠Q
AB/PQ = BD/QM (proved above)
∴ ΔABD ∼ ΔPQM (by SAS similarity)

From similarity: AB/PQ = AD/PM
Hence proved.

✅ Solution

Step 1: Use property of parallelogram

In a parallelogram, diagonals bisect each other.
Let diagonals AC and BD intersect at O.
Then O is midpoint of both AC and BD.

Step 2: Find midpoint of AC

Midpoint of AC = \(\left(\frac{-2+4}{2}, \frac{1+b}{2}\right)\) = \(\left(1, \frac{1+b}{2}\right)\)

Step 3: Find midpoint of BD

Midpoint of BD = \(\left(\frac{a+1}{2}, \frac{0+2}{2}\right)\) = \(\left(\frac{a+1}{2}, 1\right)\)

Step 4: Equate midpoints

Since both represent the same point O:
1 = (a+1)/2 ⇒ a+1 = 2 ⇒ a = 1
(1+b)/2 = 1 ⇒ 1+b = 2 ⇒ b = 1

Answer: a = 1, b = 1

✅ Solution

Step 1: Let ratio be k:1

Let point R(x, 0) divide PQ in ratio k:1
Using section formula:
x-coordinate: x = \(\frac{k×3 + 1×2}{k+1}\) = \(\frac{3k+2}{k+1}\)
y-coordinate: 0 = \(\frac{k×7 + 1×(-2)}{k+1}\) = \(\frac{7k-2}{k+1}\)

Step 2: Solve for k using y-coordinate

0 = \(\frac{7k-2}{k+1}\)
7k – 2 = 0
7k = 2
k = 2/7

Step 3: Find x-coordinate

x = \(\frac{3(2/7)+2}{(2/7)+1}\) = \(\frac{6/7+2}{2/7+1}\) = \(\frac{6/7+14/7}{2/7+7/7}\)
= \(\frac{20/7}{9/7}\) = 20/9

Answer: Ratio = 2:7, Point = (20/9, 0)

✅ Solution

Step 1: Understand trisection

Trisection means dividing into 3 equal parts.
P divides AB in ratio 1:2 (since P is nearer to A)
Q divides AB in ratio 2:1 (Q is nearer to B)

Step 2: Find coordinates of P (ratio 1:2)

P divides AB in ratio 1:2
P = \(\left(\frac{1×(-7)+2×2}{1+2}, \frac{1×4+2×(-2)}{1+2}\right)\)
= \(\left(\frac{-7+4}{3}, \frac{4-4}{3}\right)\) = \(\left(\frac{-3}{3}, \frac{0}{3}\right)\) = (-1, 0)

Step 3: Find coordinates of Q (ratio 2:1)

Q divides AB in ratio 2:1
Q = \(\left(\frac{2×(-7)+1×2}{2+1}, \frac{2×4+1×(-2)}{2+1}\right)\)
= \(\left(\frac{-14+2}{3}, \frac{8-2}{3}\right)\) = \(\left(\frac{-12}{3}, \frac{6}{3}\right)\) = (-4, 2)

Answer: P(-1, 0), Q(-4, 2)

✅ Solution

Step 1: Calculate distances

Let A(3,0), B(6,4), C(-1,3)

AB = √[(6-3)² + (4-0)²] = √(3² + 4²) = √(9+16) = √25 = 5
BC = √[(-1-6)² + (3-4)²] = √((-7)² + (-1)²) = √(49+1) = √50
AC = √[(-1-3)² + (3-0)²] = √((-4)² + 3²) = √(16+9) = √25 = 5

Step 2: Check for isosceles triangle

AB = AC = 5
BC = √50
∴ Two sides are equal ⇒ Triangle is isosceles.

Step 3: Check for right angle

Check Pythagoras theorem:
AB² + AC² = 5² + 5² = 25 + 25 = 50
BC² = (√50)² = 50
∴ AB² + AC² = BC²

Hence, triangle is right angled at A (since sides containing right angle are equal).

Conclusion: The points form a right angled isosceles triangle.

✅ Solution

Step 1: Understand given condition

AP = (3/7)AB
This means P divides AB in ratio 3:4
Because: AP = 3/7 AB

⇒ AP/AB = 3/7

⇒ AP:PB = 3:(7-3) = 3:4

Step 2: Apply section formula

A(-2,-2), B(2,-4), ratio = 3:4
P = \(\left(\frac{3×2 + 4×(-2)}{3+4}, \frac{3×(-4) + 4×(-2)}{3+4}\right)\)

Step 3: Calculate coordinates

x-coordinate = \(\frac{6 – 8}{7}\) = \(\frac{-2}{7}\)
y-coordinate = \(\frac{-12 – 8}{7}\) = \(\frac{-20}{7}\)

Answer: P\(\left(\frac{-2}{7}, \frac{-20}{7}\right)\)