Question

In the given figure(not drawn to scale), BC is parallel to EF, CD is parallel to FG, AE : EB = 2 : 3, ∠BAD = 70°, ∠ACB = 105°, ∠ADC = 40° and AC is bisector of ∠BAD.

1. Prove ΔAEF ~ ΔAGF 
2. Find: 

1. AG : AD 
2. area of ΔACB : area ΔACD 
3. area of quadrilateral ABCD : area of ΔАСВ.

Answer 1

Given:

BC || EF, CD || FG.

AE : EB = 2 : 3.

∠BAD = 70°, ∠ACB = 105°, ∠ADC = 40°.

AC is the bisector of ∠BAD, so ∠BAC = ∠CAD = 35°.

Step-wise calculation:

1. From AC bisecting ∠BAD:

∠BAC = ∠CAD

= `70^circ/2`

= 35°

2. In ΔABC:

∠ABC = 180° – ∠BAC – ∠ACB

= 180° – 35° – 105°

= 40°

3. Using the parallels:

EF || BC

⇒ ∠AEF = ∠ABC = 40°

FG || CD

⇒ ∠AGF = ∠ADC = 40°

At A, ∠EAF = ∠GAF = 35°, each is the angle between `(AE)/(AG)` and AC.

Hence, ΔAEF and ΔAGF have two equal angles 35° and 40° and so are similar AA. This proves part (a).

4. For AG : AD:

EF || BC

⇒ ΔAEF ∼ ΔABC 

So, `(AF)/(AC) = (AE)/(AB)`.

AE : EB = 2 : 3

⇒ `(AE)/(AB) = 2/(2 + 3) = 2/5`

So, `(AF)/(AC) = 2/5`.

From similarity in step 3 (ΔAGF ∼ ΔADC),

`(AG)/(AD) = (AF)/(AC) = 2/5` 

Therefore, AG : AD = 2 : 5.

5. For area(ΔACB) : area(ΔACD):

In ΔACD:

∠CAD = 35°

∠ADC = 40°

So, ∠ACD = 180° – 35° – 40° = 105°.

Thus, ΔACB and ΔACD have angles (35°, 105°, 40°) in common and they share side AC.

By ASA they are congruent, so their areas are equal.

Therefore, area(ΔACB) : area(ΔACD) = 1 : 1.

6. For area(quadrilateral ABCD) : area(ΔACB):

Quadrilateral ABCD is the union of ΔACB and ΔACD, which have equal area.

So, area(ABCD) = 2 × area(ΔACB). 

Therefore, area(ABCD) : area(ΔACB) = 2 : 1.